OK8INDERS 

iitli  First  Sirt-c 

N   JOSC,  CAL. 


^,  G,  :D|VlliH, 


^,  G.  SlVlllt^ 


Safe  Building, 


Safe  BaiLDiNG 


A  TREATISE 

GIVING   IN    THE   SIMPLEST   FORMS   POSSIBLE 

THE    PRACTICAL   AND    THEORETICAL   RULES 
AND  FORMUL/E  USED  IN  THE   CON- 
STRUCTION OF  BUILDINGS 


BY 

LOUIS   DE  COPPET  BERG,  F.  A.  I.  A. 

ASSOCIATE  AMERICAN   SOCIETY   OF   CIVIL  ENGINEERS 


VOLUME   ONE 


SECOND  EDITION,  REVISED 


BOSTON 

TICKNOR     AND     COMPANY 

2U  Cremont  .Street 
1890 


Copyright,  iS86,  1S87,  18SS,  and  1889,  by 
TICKNOR  AND   COMPANY. 


All  rizhts  reserved. 


S.  J.  Parkhill  &  Co.,  Printers,  Boston. 


To 

MY  FRIEND  AND    PARTNER, 

J.    CLEVELAND    CADY,    M.  A.,  F.  A.  L  A., 

THIS    BOOK    IS   AFFECTIONATELY    DEDICATED 

BY  THE  WRITER. 


201605'; 


•^ 


TABLE    OF   CONTENTS. 


PAGE 

LIST  OF  TABLES "*'"'  ^"^ 

LIST  OF  FORMULAE 

INTRODUCTORY  REMARKS         


IX-XU 

1 


CHAPTER  I. 
Strength  of  Materials 2-84 

CHAPTER  11. 
Foundations 85-95 

CHAPTER  III. 
Cellar  AND  Retaining  Walls 96-116 

CHAPTER  IV. 
Walls  AND  Piers       117-153 

CHAPTER  V. 
Arches 154-179 

CHAPTER  VI. 
Floor-Beams  and  Girders 180-240 

CHAPTER  VII. 
GRArmcAL  Analysis  of  Transverse  Strains.  .      .  .     241-271 


LIST   OF   TABLES. 


Table  I.  —  Distance  of  extreme  fibres,  i,  r,  a,  p-,  for 

different  sections 

Table  II.  —  Value  of  n  in  compression  formula 

Table  III.  —  'Wrinkling  strains  .... 

Table  IV.  —  Strength  and  weights  of  materials 

Table  V.  —  Strengths  and  weights  of  stones,  bricks,  and 

cements 

Table  A''I.  —  Weights  of  other  materials   . 
Table  VII.  —  Bending  moments,  shearing  and  deflection 
Table  VIII.  —  Strength  of  1"  wooden  beams 

Table  IX.  —  When  to  calculate  rupture  or  deflection 
Table  X.  —  Angles  of  friction  of  soils,  etc. 
Table  XI.  —  Values  of  p  and  slope  for  retaining  walls 
Table  XV.  —  Iron  I-beam  girders         .... 
Table  XVI.  —  A-'alue  of  y  in  formula  for  unbraced  beams 
Table  XVII.  —  Continuous  girders 
Table  XVIII.  — Trussed  beams 

Table  XII.  —  Wooden  floor-beams 
Table  XIII.  —  Wooden  girders 
Table  XIV.  —  Iron  I  beams  for  floors 
Table  XIX.  — Explaining  Tables  XX  to  XXV 
Table  XX. — List  of  iron  and  steel  I  beams 


page 

12-21 
23 
26 

37^2 


43-45 

46 

58,59 

62 

63 

09 

101 

199 

205 

218,219 

220, 221 

at  end  of  text. 


VIU  LIST    OF    TABLES. 

Table  XXI.  —  List  of  iron  and  steel  channels        .         .  at  end  of  text. 
Table  XXII.  —  List  of  iron  and  steel  even-legged  angles,  " 
Table  XXIII.  —  List  of  iron  and  steel  uneven-legged  an- 
gles          " 

Table    XXIV.  —  List  of  iron  and  steel  tees        ...  " 
Table  XXV.  —  List  of  iron  and  steel  decks,  lialf-decks, 

and  zees         .-...,  " 


LIST   OF   FORMULA. 


NO.  OF  PAGE 

FORMULA. 

1.  — Fundamental  formula  — Stress  and  Strain      .        .        .        .        •      21 

22 

2.  —  Compression,  snort  columns 

04 

3  "  long  columns •        *      " 

05 
4..— Wrinkling  strains 

28 
5. —Lateral  flexure 

30 
6. —Tension 

.      32 
7.— Shearmg,  across  gram 

30 
8.—        "  along  grain 

9__        "  at  left  reaction "'' 

10. "  at  right  reaction "^^ 

ll._Vertical  shearing  strain,  any  point  of  beam    .        .        •        .        •  3i 

^^  ((  "  "  "    cantilever        ...  35 

35 
13.  —  Longitudinal  shearing 

14. —Amount  of  left  reaction  (single  load) ^"^ 

15.  _        "  right  .        .        ,        .        ■ 

1(3___        "  left         "  ■    (two  loads) ** 

17.—        "  right  .... 

18.  — Transverse  strength,  uniform  cross-section         ....  oO 

"         upper  fibres ^^ 

"         lower      "  ... 

21. —  Central  bending  moment  on  beam,  uniform  load 
22.-        " 


19. 

20.-        ....  50 

52 

centre  load  ....  5'-^ 


23.- 
24.- 
25.- 
26.- 
27.- 
28.- 
29.- 

30.- 
31.- 
32.- 

33.- 
34.- 

35.- 
30.- 
37.  - 
38.- 

39.  ■ 

40.  ■ 
41. 
42. 
43. 
44. 
45. 
46. 
47. 
48. 
49. 
50. 
51. 


hl^T    OF    KOU.MUL.K. 

■  Bending  moment,  au}-  point,  several  loads 53 

»              '<                      "                  "            clieck  to  No.  23      .  53 

"              "         cantilever,  uniform  load 53 

'<              "               "           end  load 53 

"              "                "            any  loading 54 

-  Safe  deflection,  beams,  not  to  crack  plastering  ....  56 

-  "            "         cantilevers  or  uneven  loads  on  beams,  not  to 
crack  plastering .        .        .  5t 

-  Comparative  formula  —  beams,  strength,  different  cross-sections,  60 

«                  '<               "        deflection        "               "  61 
"                   "               "        strength,  different  lengths  and 

sections,  02 

<<                  "               "        deflection,        "                    "  62 

"                   "        columns,  different  lengtlis     ...  65 

"                   "              "          different  sections       ...  65 

-  Approximate  formula  for  plate  girders 65 

-Deflection,  cantilever,  uniform  load 66 

"                "             end  load 66 

"          beam,           uniform  load 66 

"              "                centre  load 66 

_          "              "               any  load 67 

"          cantilever,         " 67 

-Point  of  greatest  deflection  of  beam 67 

-Strain  on  arch  edge  nearest  to  line  of  pressure  .        ...  78 

"               "         farthest  from      "            79 

-  Safe  load  on  long  piles 93 

-  Amount  of  pressure  again.<t  retaining  walls,  with  higher  backing,  98 
_  "  "  "  "  with  level  backing  .  98 
-Formula  47  simplified 100 

"      48      " 100 

-Pressures  against  cellar  walls,  general  case        ....  101 

_          "               "             "            usual  case         .        .        .        :        .  101 


LIST    OF    FOKMUI.-K.  xi 

53.  —  Ke^e^■voi^•  wall.<,  fresh  water        .......  10."} 

54. —          "            "     saltwater 103 

55.  —  Retaining  wall,  backing  loaded 103 

56. —          "            "                 "             centre  of  pre.'ssiire         ,        .        .  104 

57. —          "            "     graphical  method 10.") 

58.  —  Buttressed  wall,  average  weight 107 

59.  —  Piers,  chimney  and  tower  walls,  in  inches l.'Jl 

CO.—          "           "                 "               in  feet rS-L 

61. — Flue  area  of  chimneys 1.38 

62.  —  Thickness  of  walls,  in  inches 141 

63.—          "            "           in  feet 141 

64.  —  Pressure  on  walls  from  barrels 147 

65. —Metal  bands  to  domes 174 

66.  —  Height  of  courses  in  domes 175 

67.  —  Approximate  depth  at  crown  for  arches 179 

68. —Approximate  rule  for  arches 179 

09. —  Width  of  stirrup  irons .        .  193 

70. —Thickness  of  stirrup  irons 193 

71.—          "                  "                for  wrought-iron        ....  193 

72.  —  Rectangular  beams,  transverse  strength,  uniform  load         .        .  194 

73. —          "                  "                      "                 centre  load         .        .  194 

74.—          "                  "                      "                  any  load         ...  195 

75. —          "             cantilever,            "                  uniform  load      .        .  195 

76.—          "                  "                       "                  end  load         .        .        .  195 

77.—          "                  "                      "                  any  load     .        .        .  195 

78. — Lateral  flexure  for  beams 204 

79.  —  Approximate  deflection,  iron  (also  steel)  beams  and  girders, 

uniform  load,  223 

80.—          "                   "                      "                    "             centre  load    .  224 
81.  —  Safe  length  not  to  crack  plastering,  iron  (also  steel)  beams  and 

girders,  uniform  load,  224 

82. —          "                   •'                   "                  "                   centre  load   .  224 


LIST    OF    FOKMUI..K. 


8:'..  — Avera<?e  strain  on  chords,  uniform  cross-sectiou  and  load 
84.—  "  top  chord  "  "  .        . 

So. —  "  bottom  chord    "  "  .         . 

8(!. —  "  top  chord,  uniform  section,  centre  load    . 

87.—  "  bottoni  chord    "  "  .         . 

88.  —  Expansion  or  contraction  from  strain 

8'.t.  —  Detlection  of  girder,  parallel  flanges,  uniform  section 
90.—  "  "  varying  section  . 

91.  — Safe  lengtli  not  to  crack  plastering,  parallel  flanges,  varying  sec- 
tion    ............ 


92. —  Graphical 

93.— 

94.— 

95.— 

96.— 

97.— 

98.— 

m.— 


iiethod  —  moment  of  resistance     .        .        .        . 
bending  moment         .... 

cross-shearing 

deflection  witli  i  .... 

deflection  with  v 

deflection,  random  pole  distance 
thickness  of  plate  girder  flange    . 
value  of  angles  in  diminishing  flanges 


225 
226 
226 

007 

227 
227 
228 
229 

230 
245 
245 
246 
248 
248 
248 
270 
270 


SAFE    BUILDING. 


INTRODUCTORY  REMARKS. 

IN  the  articles  on  this  suhject  the  writer  proposes  to  furnish  to  any 
earnest  student  the  opportunity  to  acquire,  so  far  as  books  will 

teach,  the  knowledge  necessary  to  erect  safely  any  building.  While, 
of  course,  the  work  will  be  based  strictly  on  the  science  of  mechanics, 
all  useless  theory  will  be  avoided.  The  object  will  be  to  make  the 
articles  simply  practical.  To  fallow  any  of  the  mathematical  demon- 
strations, arithmetic  and  a  rudimentary  knowledge  of  algebra  and 
plane  geometry  will  be  sufficient. 

The  following  outline  will  probably  give  a  better  idea  of  the  work 
proposed  :  — 

First  will  come  an  introductory  chapter  on  the  "  Strength  of 
Materials."  This  chapter  will  give  the  values  of,  and  explain  briefly, 
the  different  terms  used,  such  as  strain,  stress,  factor-of-safety, 
centre  of  gravity,  neutral  axis,  moment  of  inertia,  centre  and  radius 
of  gyration,  moment  of  resistance,  and  moduli  of  elasticity  and  rup- 
ture. 

Then  will  follow  the  several  formulas  to  be  used,  with  explanations 
giving  their  applications,  viz. :  compression  in  long  and  short  columns ; 
wrinkling  strains  and  lateral  llexure  in  top  chords  of  girders  and 
beams ;  tension  and  sliearing  strains ;  transverse  strains,  including 
rupture,  deflection  and  bending  moments  in  cantilevers  and  beams ; 
parallelogram  of  forces  and  graphical  method  of  calculating  trusses 
and  arches ;  also  manner  of  obtaining  amounts  of  loads.  Accom- 
panying the  above  will  be  the  necessary  tables  used  in  calculations. 

After  this  introductory  chapter  will  follow  a  series  of  chapters, 
each  dealing  with  some  part  of  a  building,  giving  practical  advice 
and  numerous  examples  of  calculations  of  strength ;  for  instance, 
chapters  on  foundations,  walls  and  piers,  columns,  beams,  rivited 
and  other  girders,  cast-iron  lintels,  roof  and  other  trusses,  spires; 
masonry,  inverted  and  floor  arches,  corrugated  iron,  stairs,  sidewalks, 
chimneys,  etc.,  and  possibly  also  chapters  on  drainage,  plumbing, 
heatinor  and  ventilatins- 


SAFE    BUILDING. 


CHAPTER   I. 

STRENGTH    OF   MATERIALS. 


(German,  FesligTceit;  French,  Resistance  des  materiaux.) 
All  solid  bodies  or  materials  are  made  up  of  an  infinite  number  of 
atoms,  fibres  or  molecules.  These  adhere  to  each  other  and  resist 
separation  with  more  or  less  tenacity,  varying  in  different  materials. 
This  tenacity  or  tendency  of  the  fibres  to  resume  their  former  rela- 
tion to  each  other  after  the  strain  is  removed  is  called  the  elasticity 
of  the  material.  It  is  when  this  elasticity  is  overcome  that  the  fibres 
separate,  and  the  material  breaks  and  gives  way. 

There  are  to  be  considered  in  calculating  strengths  of  materials  two 
kinds  of  forces,  viz.,  the  external  (or  applied)  forces  and  the  inter- 
nal (or  resisting)  forces.  The  external  forces  are  any  kind  of  forces 
applied  to  a  material  and  tending  to  disrupt  or  force  the  fibres  apart. 
Thus  a  load  lying  apparently  perfectly  tranquil  on  a  beam  is  really 
a  very  active  force;  for  the  earth  is  constantly  attracting  the  load, 
which  tends  to  force  its  way  downwards  by  gravitation  and  push 
aside  the  fibres  of  the  beam  under  it.  These  latter,  however,  resist 
separation  from  each  other,  and  the  amount  of  the  elasticity  of  all 
these  fibres  being  greater  than  the  attraction  of  the  earth,  the  load 
is  unable  to  force  its  way  downwards  and  remains  apparently  at  rest. 
Strain.  The  amount  of  this  tendency  to  disrupt  the  fibres 
(produced  by  the  external  forces)  at  any  point  is  called  the  "  strain  " 
at  that  point. 

Stress.         The  amount  of  the  resistance  against  disruption  of 
the  fibres  at  such  point  is  called  the  "  stress"  at  that  point. 

External  (or  applied)  forces,  then,  produce  strains.  Internal  (or 
resisting)  forces  produce  stresses. 

This  difference  must  be  well  understood  and  constantly  borne  in 
mind,  as  strains  and  stresses  are  the  opposing  forces  in  the  battle  of 
all  materials  against  their  destruction. 

When  the  strain  at  every  point  of  the  material  just 
stress,  equals  the  stress,  the  material  remams  in  equilibrium. 
The  o-reatest  stress,  at  any  point  of  a  material  that  it  is  capable  of  ex- 
erting is  the  ultimate  stress  (that  is,  the  ultimate  strength  of  resist- 
ance) at  that  point.  Were  the  strain  to  exactly  equal  that  ultimate 
stress,  the  material,  though  on  the  point  of  breaking,  would  still  be 


8TKENGTII    OF    MAIKKIALS.  S 

safe  thcoreticallv.     But  it  is  impossible  for  us  to  oal- 

,   ,  1      1 '       T>     •  1  ,  Factor-of- 

culate  so  closely.      Besides  we  can  never  determine  Safety. 

accurately  the  actual  ultimate  stress,  for  different  jiieces  of  the  same 
material  vary  in  practice  very  greatly,  as  has  been  often  proved  by 
experiment.  Therefore  the  actual  ultimate  stress  might  be  very  much 
less  than  that  calculated. 

Again,  it  is  impossible  to  calculate  the  exact  strain  that  will  alwavs 
take  place  at  a  certain  point ;  the  applied  forces  or  some  other  con- 
ditions might  vary.  Therefore,  to  provide  for  all  possible  emergen- 
cies, we  must  make  our  material  strong  enough  to  be  surely  safe ;  that 
is,  we  must  calculate  (allow)  for  a  considerably  greater  ultimate  stress 
at  every  point  than  there  is  ever  likely  to  be  strain  at  that  point. 

The  amount  of  extra  allowance  of  stress  varies  greatly,  according 
to  circumstances  and  material.  The  number  of  times  that  we  calcu- 
late the  ultimate  stress  to  be  greater  than  the  strain  is  called  the  fac- 
tor-of-safety  (that  is,  the  ratio  between  stress  and  strain). 

If  the  elasticity  of  different  pieces  of  a  given  material  is  practi- 
cally uniform,  and  if  we  can  calculate  the  strain  very  closely  in  a 
given  case,  and  further,  if  this  strain  is  not  apt  to  ever  vary  greatly, 
or  the  material  to  decay  or  deteriorate,  we  can  of  course  take  a  low 
or  small  factor-of-safety ;  that  is,  the  ultimate  stress  need  not  exceed 
many  times  the  probable  greatest  strain. 

On  the  other  hand,  if  the  elasticity  of  different  pieces  of  a  given 
material  is  very  apt' to  vary  greatly,  or  if  we  cannot  calculate  the 
strain  very  closely,  or  if  the  strain  is  apt  to  vary  greatly  at  times,  or 
the  material  is  apt  to  decay  or  to  deteriorate,  we  must  take  a  very 
high  or  large  factor-of-safety,  that  is,  the  ultimate  stress  must 
exceed  many  times  the  probable  greatest  strain. 

Factors-of-safety  are  entirely  a  matter  of  practice,  experience,  and 
circumstances.     In  general,  we  might  use  for  stationary  loads : 
A  factor  of  safety  of  3  to  4  for  wrought-iron  and  steel, 
"  "  "     6  for  cast-iron, 

"  "  "     4  to  10  for  wood, 

"  "  "10  for  brick  and  stone. 

For  moving-loads,  such  as  people  dancing,  machinery  vibratin"-, 
dumping  of  heavy  loads,  etc,  the  factor-of-safety  should  be  one- 
half  larger,  or  if  the  shocks  are  often  repeated  and  severe,  at  least 
double  of  the  above  amounts.  Where  the  constants  to  be  used  in 
formuliB  are  of  doubtful  authority  (as  is  the  case  with  most  of  them 
for  woods  and  stones),  the  factor-of-safety  chosen  should  be  the  hi"-h- 
est  one. 


SAFE    BUILPING. 


In  building-materials  we  meet  with  four  kind  of  strains,  and,  of 
course,  with  the  four  correspondiag  sti-esses  resisting  them,  viz.:  — 


Compression,  or  crushing  strains. 
Tension,  or  pulling  strains, 
Shearing,  or  sliding  strains,  and 
Transverse,  or  cross-breaking  strains. 

STRESSES. 

The  resistance  to  Compression,  or  crushing-stress. 

The  resistance  to  Tension,  or  pulling-stress. 

The  resistance  to  Shearing,  or  sliding-stress,  and 

The  resistance  to  Transverse  strains,  or  cross-breaking  stress. 

Materials  yield  to  Compression  in  three  different  ways  :  — 

1.  By  direct  crushing  or  crumbling  of  the  material,  or 

2.  By  gradual  bending  of  the  piece  sideways  and  ultimate  rupture,  or 

3.  By  buckling  or  wrinkling  (corrugating)  of  the  material  length- 
wise. 

^laterials  yield  to  Tension, 

1.  By  gradually  elongating  (stretching),  thereby  reducing  the  size 
of  the  cross-section,  and  then, 

2.  By  direct  tearing  apart. 

Materials  yield  to  Shearing  by  the  fibres  sliding  past  each  other  in 
two  different  ways,  either 

1.  Across  the  grain,  or 

2.  Lengthwise  of  the  grain. 
Materials  yield  to  Transverse  strains, 

1.  By  deflecting  or  bending  down  under  the  load,  and  (when  thia 
passes  beyond  the  limit  of  elasticity), 

2.  Bv  breaking  across  ti-ansversely. 

In  calculating  strains  and  stresses,  there  are  certain  rules,  expres- 
sions, and  formulae  which  it  is  necessary  for  the  student  to  under- 
stand or  know,  and  which  will  be  here  given  without  attempting  elab- 
orate explanations  or  proofs.  For  the  sake  of  clearness  and  simplic- 
ity, it  is  essential  that  in  all  formulae  the  same  letters  should  alwaijs 
represent  the  same  value  or  meaning  ;  this  will  enable  the  student  to 
read  every  formula  off-hand,  without  the  necessity  of  an  explanatory 
key  to  each  one.  The  writer  has  further  made  it  a  habit  to  express, 
in  all  cases,  his  formuUe.  in  pounds  and  inches  (rarely  using  tons  or 
feet)  ;  this  Avill  frequently  make  the  calculation  a  little  more  elabo- 


GLOSSAKY   OV    SYMBOLS.  5 

rate,  but  it  will  be  found  to  greatly  simplify  the  formulae,  and  to  make 
their  understanding  and  retention  more  easy. 

In  the  following  articles,  then,  a  capital  letter,  if  it  were  used, 
would  invariably  express  a  quantity  (respectivel}'),  either  in  tons  or 
feet,  while  a  small  letter  invuriabl>j  expresses  a  quantity  (respec- 
tively), either  in  pounds  or  inches. 

The  following  letters,  in  all  cases,  will  be  found  to  express  the  same 
meaning,  unless  distincthj  otherwise  staled,  viz. :  — 
a    signifies  area,  in  square  inches. 
b  "       breadth,  in  inches. 

c  "       constant  for  ultimate  resistance  to  compression,  in  pounds, 

per  square  inch.     (See  Tables  IV  and  V.) 
d  signifies  depth,  in  inches. 
e  "       constant  for  modulus  of  elasticity,  in  pounds-inch,  that  is, 

pounds  per  square  inch.     (See  Table  IV.) 
f  "       factor  of  safety. 

g  "        constant  for  ultimate  resistance    to  shearing,  per  square 

inch,  across  the  grain.     (See  Tables  IV  and  V.) 
g^  "        constant   for  ultimate  resistance  to  shearing,  per  scjuare 

inch,  lengthwise  of  the  grain.     (See  Table  IV.) 
h  "       height,  in  inches. 

i  "        moment  of  inertia,  in  inches.     (See  Table  I.) 

k  "        ultimate  modulus  of  rupture,  in  pounds,  per  square  inch. 

(See  Tables  IV  and  V.) 
I  "       length,  in  inches. 

771         "       moment  or  bending  moment,  in  pounds-inch.  (See  Table  IX.) 
n  "       constant  in  Rankine's   formula  for  compression  of  long 

pillars.     (See  Table  II.) 
0  "        the  centre. 

p  "        the  amount  of  the  left-hand  re-action  (or  support)  of  beams, 

in  pounds. 
q  "        the  amount  of   the  right-hand  re-action  (or  support)  of 

beams,  in  pounds. 
r  "        moment  of  resistance,  in  inches.     (See  Table  I.) 

s  "        sti'ain,  in  pounds. 

t  "       constant  for  ultimate  resistance  to  tension,  in  pounds,  per 

square  inch.     (See  Tables  IV  and  V.) 
u  "        uniform  load,  in  pounds. 

V  "       stress,  in  pounds. 

to  "        load  at  centre,  in  pounds. 

X,  y,  and  z  signify  unknown  quantities,  either  in  pounds  or  inches. 


€  SAFE   BUILDING. 

(3    signifies  total  dejlectlon,  in  inches. 
p2         «       square  of  the  radius  of  gyration,  in  inches. 
jj  "        diameter,  in  inches. 

f  "       radius,  in  inches. 

ff  =  3.14159,  or,  say,  3.1-7  signifies  the  ratio  of  the  circun}ference  and 
diameter  of  a  circle. 

If  there  are  more  than  one  of  each  kind,  the  second,  third,  etc.,  are 
indicated  with  Roman  numerals,  as  for  instance,  a,  a^,  a„,  a„„  etc.,  or 
b,  h„  6,„  6„„  etc. 

In  taking  moments,  or  bending  mo.ments,  strains,  stresses,  etc.,  to 
signify  at  what  point  they  are  taken,  the  letter  signifying  that  point 
is  added,  as  for  instance  :  — 
m  signifies  moment  or  bending  moment  at  centre. 
7,)^      "  "  "  "  point  A. 

m^      "  "  "  "  point  B. 

m^      "  "  "  "  pomf  X. 

s         "         strain  at  centre, 
s^        "  "     poiiit  B. 

Sx        "  "         "      X. 

V         "         stress  at  centre. 
I'p       "  "    point  D. 

v^        «  "  X 

tv    signifies  load  at  centime. 
u\  "         "     "    point  A, 

CENTKE   OF   GRAVITY. 

(German,  Scliwerpunkt;  French,  Centre  de  gravite.) 
The  centre  of  gravity  of  a  figure,  or  body,  is  that  centre  of 

point  upon  which  the  figure,  or  body,  will  balance  Gravity. 

itself  in  whatever  position  the  figure  or  body  may  be  placed,  provided 

no  other  force  than  gravity  acts  upon  the  figure  or  body. 

To  find  the  centre  of  gravity  of  a  plane  figure,  find  two  neutral 

axes,  in  different  directions,  and  their  point  of  intersection  will  be 

the  centre  of  gravity  required. 

NEUTRAL    AXIS. 

(German,  Neutrale  Achse;  French,  Axe  neutre.') 
The  neutral  axis  of  a  body,  or  figure,  is  an  imagin-  Neutral  Axis. 
ary  line  upon  which  the  body,  or  figure,  will  always  balance,  provided 
the  body,  or  figure,  is  acted  on  by  no  other  force  than  gravity.     The 
neutral  axis  always  passes  through  the  centre  of  gravity,  and  may  run 
in  anv  direction.     In  calculating  transverse  strains,  the  neutral  axis 


NEUTRAL    AXIS.  7 

designates  an  imaginary  line  of  the  body,  or  of  the  cross-section  of  the 
body,  at  which  the  forces  of  compression  and  tension  meet.  The  strain 
on  the  fibres  at  the  neutral  axis  is  always  naught.  Extreme  Fibres. 
On  the  upper  side  of  the  neutral  axis  the  fibres  are  compressed,  while 
those  on  the  lower  side  are  elongated.  The  amount  of  compression 
or  elongation  of  the  fibres  increases  directly  as  their  distance  from 
the  neutral  axis ;  the  greatest  strain,  therefore,  being  in  the  fibres 
along  the  upper  and  lower  edges,  these  being  farthest  from  the 
neutral  axis,  and  tlierefore  called  the  extreme  fibres.  It  is  necessary 
to  calculate  only  the  ultimate  resistance  of  these  extreme  fibres,  as,  if 
they  will  stand  the  strain,  certainly  all  the  other  fibres  will,  they  all 
being  nearer  the  neutral  axis,  and  consequently  less  strained. 
Where  the  ultimate  resistances  to  compression  and  tension  of  a 
material  vary  greatly,  it  is  necessary  to  so  design  the  cross-section  of 
the  body,  that  the  "  extreme  fibres "  (farthest  edge)  on  the  side 
offering  the  weakest  resistance,  shall  be  nearer  to  the  neutral  axis 
than  the  "  extreme  fibres "  (farthest  edge),  on  the  side  offering 
the  greatest  resistance,  the  distance  of  the  "  extreme  fibres  "  from 
the  neutral  axis  being  on  each  side  in  direct  proportion  to  their 
respective  capacities  for  resistance.  Thus,  in  cast-iron  the  resistance 
of  the  fibres  to  compression  is  about  six  times  greater  than  their 
resistance  to  tension ;  we  must  therefore  so  design  the  cross-section, 
that  the  distance  of  the  neutral  axis  from  the  top-edge  will  be  six- 
sevenths  of  the  total  depth,  and  its  distance  from  the  lower  edge 
one-seventh  of  the  total  depth. 

To  find  the  neutral  axis  of  any  plane-figure,  some  writers  recom- 

/^7\ mt-^nd  cutting,  in  ^^^  ^^  „„j, 

7  ~'Y~    stiff  card-board.       Neutral  Axis. 

T  I        aduplicateof  the  figure  (of  which 

-ts X — \        the  neutral  axis  is  sought),  then 

— — Nto  experiment  until  it  balances 

)H^  -n        J  on  the  edge  of  a  knife,  the  line 

c\       ^m     ^  -- '-^■^ — ■ f    Oil 

7     Ltj^r — :i2L^   ij  on   which  it  balances   beinsr,  of 

L L                dir  I  '^ 

y. I     r ^         '^  \i    [  \     -y  course,   the  neutral  axis.     This 

Fig.  I.  is  an  awkward  and  unscientific 

method  of  procedure,  though  there  may  be  some  cases  where  it  will 

recommend  itself  as  saving  time  and  trouble. 

The  following  general  formula,  however,  covers  every  case :   To 

find  the  neutral  axis  M-N  in  any  desired  direction,  draw  a  line 

X-Y  at  random,  but  parallel  to  the  desired  direction.     Divide  the 

figure  into  any  number  of  simple  figures,  of  which  the  areas  and  cen^ 


8  SAKE    BUILDING. 

tres  of  gravity  can  be  readil}-  found,  then  the  distance  of  the  neutral 
axis  M-N  from  the  line  X-  Y  will  be  equal  to  the  sum  of  the  productn  ^ 
of  each  of  the  small  areas,  multiplied  by  the  distance  of  the  centre  of 
gravity  of  each  area  from  X-Y,  the  whole  to  be  divided  by  the  entire 
area  of  the  whole  figure.     An  Example  will  make  this  more  clear. 

Find  the  horizontal  neutral  axis  of  the  tross-section  of  a  deck-beam, 
standing  vertically  on  its  bottom-Jlange. 

Draw  a  line  {X-Y)  horizontally  (Fig.  1),  then  let  (/„  J,„  c/,„,  rep 
resent  the  respective  distances  from  X-Y  oi  the  centres  of  gravity 
of  the  small  subdivided  simple  areas  a„  a„,  a,„,  then  let  a  stand  for 
the  whole  area  of  section,  that  is  :  — 

«.  +  «,i  +  «m  =a, 
then  the  required  distance  (d)  of  the  neutral  axis  M-iVfrom  X-Y, 
will  be 

7  a,  d,  4-  a,,  d,,  -\-  a,,,  d,,, 
a 
To  find  the  centre  of  gravity  of  the  figure,  we  might  find  another 
neutral  axis,  but  in  a  different  direction,  the  point  of  intersection  of 
the  two  being  the  required  centre  of  gravity.  But  as  the  figure  is 
uniform,  we  readily  see  that  the  centre  of  gravity  of  the  whole  figure 
must  be  half-way  between  points  A  and  B. 

Centres  of  '^^^  centre  of  gravity  of  a  circle  is  always  its  cen- 

Cravity.  tre.  The  centre  of  gravity  of  a  parallelogram  is  al- 
ways the  point  of  intersection  of  its  two  diagonals.  The  centre  of 
orravity  of  a  triangle  is  found  by  bi-secting  two  sides,  and  connecting 
these  points  each  with  its  respective  opposite  apex  of  the  triangle, 
the  point  of  intersection  of  the  two  lines  being  the  required  centre  of 
gravitv,  and  which  is  always  at  a  distance  from  each  base  equal  to 
one-third  of  the  respective  height  of  the  triangle.  Any  line  drawn 
through  either  centre  of  gravity  is  a  neutral  axis. 

MOMENT   OF    INERTIA. 

(German,  Trdgheitsmoment ;  French,  Moment  d'enertie,  or   Moment 
de  giration.) 

Moment  of  iner-  "^^^^  moment  of  inertia,  sometimes  called  the  mo- 
tia.  cseeTabiei.)  ment  of  gyration,  is  the  formula  representing  the 
inactivity  (or  state  of  rest)  of  any  body  rotating  around  any  axis. 
The  reason  of  the  connection  of  this  formula  with  the  calculation  of 
strains  and  the  manner  of  obtaining  it  cannot  be  gone  into  here,  as  it 
would  be  quite  beyond  the  scope  of  these  articles.  The  moment  of 
'  If  line  X-Y  is  inside  of  (bisects)  figure,  take  sum  of  products  on  one  side 
only  and  deduct  sum  of  products  on  other  side. 


TIIK    CKNTRK    OF    GYUATIOX    AXI)    KADIUS    OK    GYKATKJX.  0 

inertia  of  any  body  or  figuro  is  llie  sum  of  the  products  of  each  par- 
ticle of  the  body  or  figure  multiplied  by  the  S(]uare  of  its  distance 
from  the  axis  around  which  the  body  or  figure  is  rotatin"-. 

A  table  of  moments  of  inertia,  of  various  sections,  -will  be  "iven 
later  on  and  will  be  all  the  student  will  need  for  practical  purposes. 

THE  CENTRE  OF  GYRATION  AND  RADIUS  OF  GYRATION. 

(German,    Tragheitsmiltelpunkt ;  French   Centre  de  fjiration.) 
The  centre  of  gyration  "is  that  point  at  which,  if 
the  whole  mass  of  a  body  rotating  around  an  axis  us^'of'^Cyratfon" 
or  point  of  suspension  were  collected,  a  given  force        (^ce  Table  i.) 
applied  would  produce  the  same  angular  velocity  as  it  would  if  ap- 
plied at  the  same  point  to  the  body  itself."    The  distance  of  this  cen- 
tre of  gyration  from  the  axis  of  rotation  is  called  the  radius  of  gyra- 
tion (German,  TragheilsJialbmesser ;  French,  Rayon  de  giralion)  ;  this 
latter  is  used  in  the  calculation  of  strains,  and  is  found  by  dividin^r 
the  moment  of  inertia  of  the  body  by  the  area  or  mass  of  the  body, 
and  extracting  the  square  root  of  the  quotient,   or, 

9=Sl^,    or 

s   —     a- 
A  table  will  be  given,  later  on,  of  the  "  squares  of  the  radius  of  o-yra- 
tion"  (German,  Quadrat  des  TragheUslialhmessers ;  French,  Carri  du 
rayon  de  giralion). 

THE    MOMENT    OF    Kp:SISTANCE. 

(German,  Widerstandsmoment ;  French,  Moment  de  resistance). 

The  moment  of  resistance  of  any  fibre  of  a  body, 
revolving  around  an  axis,  is  equal  to  the  moment  of  sistamfe'!*°ee'T'J 
inertia  of  the  whole  body,  divided  by  the  distance  of  ^^^  ^'^ 
said  fibre  from  the  (neutral)  axis,  around  which  the  body  is  revolvin"-. 

A  table  of  moments  of  resistance  will  be  given  later  on. 

MODULUS   OF   ELASTICITY. 

(German,  Elasticitats  modulus,  French,  Module d'elasticite). 
The  modulus  of  elasticity  of  a  given  material  is 
the  force  required  to  elongate  a  piece  of  the  mate-  Elasticity."sceTi^ 
rial   (whose  area  of  cross-section  is  equal  to  one  '"^  ^^'•* 
square  inch)  through  space  a  distance  equal  to  its  primary  length. 
Thus,  if  a  bar  of  iron,  twelve  inches  long,  and  of  one  square  inch 
area  of   cross-section,  could   be   made   so  elastic  as   to   stretch    to 


10  SAFE    BUILDING. 

twice  its  length,  the  force  recjuired  to  stretch  it  until  it  were  twenty- 
four  inches  long  would  be  its  modulus  of  elasticity  in  weight  per 
scjuare  inch. 

MODULUS    OF    RUPTURE. 

(German  Bruchcoefficient ;  French,  Module  de  rupture.^ 

It  has  been  found  by  actual  tests  that  though  the 

Modulus      of     ,.^  -,  r  .    1  T  • 

Rupture.    (SeeTa-  diiierent  ubres  of  materials  under  transverse  strains 
"^^      ""     ■'  are  either  in  compression  or  tension,  the  ultimate 

resistance  of  the  "extreme  fibres"  neither  entirely  agrees  with  their 
ultimate  resistance  to  compression  nor  tension.  Attempts  have  been 
made  to  account  for  this  in  many  different  ways ;  but  the  fact  re- 
mains. It  is  usual,  therefore,  where  the  cross-section  of  the  material 
is  uniform  above  and  below  the  neutral  axis,  to  use  a  constant  derived 
from  actual  tests  of  each  material,  and  this  constant  (which  should 
always  be  applied  to  the  "  extreme  fibres,"  i.  e.,  those  along  upper  or 
lower  edge)  is  called  the  modulus  of  rupture,  and  is  usually  expressed 
in  pounds  per  square  inch. 

TO   FIND   THE   MOMENT   OF   INERTIA   OF   ANY  CROSS-SECTION. 

Howtofindmo-  Divide  the  cross-section  into  simple  parts,  and 
SflTnycross-Sc^fi^'i  t'^^  moment  of  inertia  of  each  simple  part 
tion.  around  its  own  neutral  axis  (parallel  to  main  neu- 

tral axis) ;  then,  if  we  call  the  moment  of  inertia  of  the  whole 
cross-section  i,  and  that  of  each  part  i„  t,„  t,„,  z,„„  etc.,  and,  fur- 
ther, it  we  call  the  area  of  each  part  a„  a,„  a,„,  a,„„  etc.,  and  the 
distance  of  the  centre  of  gravity  of  each  part  from  the  main  neutral 
axis  of  the  whole  cross-section,  </„  J,„  </,„.  f7„„,  etc.,  we  have:  — 
?=(f/,2a,+'.)+('A,^a..+i„)+(^,,,'an.-f*'...)+(fA.n''a„,,+''wn)+>etc. 
Referring  back  to  Figure  1,  we  should  have  for  Part  I :  — 

t,  =  11 1,*.       (See  Table  I.,  column  8.) 

For  Part  II  we  should  have :  — 

'" — rr 

And  for  Part  III:  — 

'■"—     12 
For  the  distances  of  individual  centres  of  gravity  from  main  centre 
of  gravity  we  should  have  for  Part  I :  d^-d. 
For  Part  II:  d,,-d. 
And  for  Part  III:  d-d,„. 


TO  FIND  THE  MOMENT  OF  INERTIA  OF  ANY  CROSS-SECTION.     11 

Therefore  the  moment  of  inertia,  i,  of  the  wliole  deck-beam  would 


be:  — 


But  a,=Bx,^ 

Further,  a„  =  &„  ft„» 

And  a„.  =  6,„  A„„  which,  inserted  above,  gives  for 

The  following  table  (I)  gives  the  values  for  the  moment  of  inertia 
(i),  moment  of  resistance  (r),  area  (a),  square  of  radius  of  gyration 
(0*),  etc.,  for  nearly  every  cross-section  likely  to  be  used  in  building. 
Those  not  given  can  be  found  from  Table  I  by  dividing  the  cross- 
section  into  several  simpler  parts,  for  which  examples  can  be  found 
in  the  table.  Note,  that  it  makes  a  great  difference  whether  the  neu- 
tral axis  is  located  through  the  centre  of  gravity  (of  the  part),  or 
elsewhere.  When  making  calculations  we  must,  of  course,  insert  in 
the  different  formulae  in  place  of  i,  r,  a,  Q^  their  values  (for  the  re- 
spective cross-section),  as  given  in  Table  I. 


12 


SAFE    BUILDING. 


TABLE  I. 

DISTANCE  OF  EXTREME  FIBRES,  MOMENTS  OF  INERTIA  AND  RESISTANCE, 
SQUARE  OF  RADIUS  OF  GYRATION,  AND  AREAS  OF  DIFFERENT  SHAPES 
OF  CROSS-SECTIONS. 


Kumber  aud  Form  of 
Section. 


5  •    % 


•*  W 


CO 

o  C 


1 

d-4 


TFWTcU 


^^ 


'■K<^x- 


■;-N 


12 


M3 
12 


12 


&f/3-i/Z,3 


12 


11 

r4^ 


^3 

6 


6 


6d 


hd^-h,d,^ 


Gd 


11 


rZ2 


?/(/ 


(/2-f/2 


bd-b,d. 


22  . 


12 


12 


</2+^,2 


12 


+ 


TABLE    I,    CONTINUED. 


13 


:?  ?  o 

g 

V 

Cfl 

istai 

alas 
om 

o 
S 
a 

o 

3 

t> 

aS 

Number  and  Form  of 

i^   X   "   a 
5  f*  s* 

a 

■^, -r?  o 

Section. 

."  s  r'  o 

»-*» 

M» 

S  : 

•■'S' 

a 

§  •    !z! 

a 

?  fe!  ? 

so' 

■  S- 

i' 

iMTic^ 


"3-ct, 


IMiy 


'M-;l--l-I\J 


l>i 

? 

c; 

^ 
" 

C 

.• 

C5 

"co 

"r 
"1 

-L 
1" 

r 

"eo 

14 


SAFE    BUILDI^XG. 


Number  and  Form  of 
Section. 


b  — -4j 


+1 


+ 


+ 


!  " 

'  Si. 


''r 

4- 


+  r 


+  + 


TABLK    I,    CONTINUKD. 


Id 


Number  and  Form  of 
Section. 


2    S.  w 


17 


MttN,-i 


l^-^J 


18 


r   I^iticed   I 


kthced 


+ 


+ 


+ 


+ 


+ 


+ 


Si     »5 


—  t 


+1^ 


T^lA 


12 


16 


SAFE    BUILDING. 


Number  and  Form  of 
Section. 


p  S 


§  2. 


fe:  ? 


■^  w 


o  ® 


+ 


+  ' 


a, 
+ 


"K  *^ 


+ 


I     ® 


+ 


+  , 


t? 


TAULE    I,    CONTINUED. 


17 


Number  and  Form  of 

Distance 

tral  axis 

from   ex 

br 

o 

B 
» 

o 

S 

»  a 

c  a 

o    n- 

CO 

o  3 

Section. 

of   Neu- 

M N 

treme    fl- 
ea. 

M 

D 

a 

^1 

+ 


I        o 


+     i 

(5-         I 


+ 


wo 
_i_     "O 

C^        hi 
I  IB 


+ 


+ 


+ 


+ 


+ 


t 
+ 


-f 


+ 


18 


SAFE    BUILDING. 


Kuniber  and  Form  of 
Section. 


^    X    a=    » 

•    2        '^ 
B 
o 


C     3 


-i 

Q 

A 

?^ 

,^ 

:^ 

r 

r 

'I 

1 

M 

M 

t" 

X 

+ 

r- 

C5 

x\ 

+ 

;+ 


+ 


+ 


:+ 


+ 


+ 


:  + 


+: 


+ 


TABLE    I,    CONTINUED. 


19 


n   if  O 

S 

■> 

CO 

O    »    X 

o 

o 

Number  ami  Form  of 

^   X     "     » 

S   2-  f 

3 
S 

?  o 

> 

Section. 

B  : 

a 

r^  ^* 

a 

o  :::r 

»   •    !^ 

a 

:    o 

Vi' 

— * 

»  !z!  c 

S' 

"T 

5 

2s 


M- 


M--^ 


M- 


-t-N 


M — 


Lower 
Fibres. 

iL 

3 
Upper 
Fibres. 

_2^ 

3d 


+ 


3 


12 


MS 
i 


36 


4- 


12 


M^ 
4 


Lower 
Fibres. 

12 

TTpper 

Fibres. 

M^ 

24 


hd 


f/2 


&(Z 


?;./ 


9 


+ 


+ 


3 


6 


18 


20 


SAFE    BUILDING. 


Number  and  Frirni  of 
Section. 


B     !o     « 


h_        b    1 
,/2"      1.4142 


mIBIn 


T 


37 


M- 


51  s 


frr      --^ 


12 


+ 


+ 


^,. 

c^'  ^ 

►^ :  <— ' 

^  •— * 

<-! 

t(^'  l— 

i4—     •— ' 

<-t 

--t 

tc\  ~£. 

rt .  "^ 

1^ 

!  ^ 

»£  r 

^'  i 

+ 


?K- 


5-W 


0.1179  h^\      Ifi 


+ 


?/2 

12 


4- 


I -J     »^ 


TABLE   I,    CONTINUED. 


21 


Number  and  Form  of 
Sectiou. 


»-»>    r^     i-H 

g 

2 

d   >=   S- 

0 

0 

o    .^ 

taiioe 

1  axis 

m    ex 

bi 

2 

, 

;4  V 

a 

3 

?    0 

> 

a 

5  ?  "  = 

B    '. 

0 

:*  w 

e 

^» 

«  1  iz: 

« 

fl 

3)  i^*   S 

P 

'^ 

CALCULATION    OF    STRAINS    AND    STRESSES. 

As  we  have  already  noticed,  the  stress  should  exceed  the  strain  as 
many  times  as  the  adopted  factor-of-safety,  or :  — 

_ — 11-^  factor-of-safety. 
Strain 

Or,  stress  =  strain  X  factor-of-safety. 

This  hokls  good  for  all  calculations,  and  can  be  expressed   by  the 

following  simple  fundamental  formula  :  —  Fundamental 

v^=.s.f  (1)  Formula- 

Where  v  =:  the  ultimate  stress  in  pounds. 

"        s  =    "    strain  in  pounds. 


22  SAFE   BUILDING. 

And  where /=  the  factor-of-safety. 

COMPRESSION. 

Compression,  In  pieces  under  compression  the  load  is  directly 
s  or  CO  um  .  j^ppjj^j  ^^  ^^xe  material.  In  short  pieces,  therefore, 
-which  cannot  give  sideways,  the  strain  will  just  equal  the  load,  or 
we  have  :  — 

s  =  w. 

Where  s  =  the  strain  in  pounds. 

And  where  ic  ^=  the  load  in  pounds. 

The  stress  will  be  equal  to  the  area  of  cross-section  of  the  piece 
being  compressed,  multiplied  by  the  amount  of  resistance  to  com- 
pression its  fibres  are  capable  of.^  This  amount  of  resistance  to 
compression  which  its  fibres  are  capable  of  is  found  by  tests,  and  is 
given  for  each  square  inch  cross-section  of  a  material.  A  table  of 
constants  for  the  resistance  to  crushing  of  different  materials  will 
be  given  later  on. 

In  all  the  formulae  these  constants  are  represented  by  the  letter  c. 

We  have,  then,  for  the  stress  of  short  pieces  under  compression  : — 
v^a.c 

Where  v  is  the  ultimate  stress  in  pounds. 

Where  a  is  the  area  of  cross-section  of  the  piece  in  inches. 

And  where  c  is  the  ultimate  resistance  to  compression  in  pounds 
per  square  inch. 

Inserting  this  value  for  v,  and  w  for  s  in  the  fundamental  formula 
(1),  we  have  for  short  pieces  under  compression,  which  cannot  yield 
sideways :  — 

a.c  =  io ./,  or  :  — 

t,=  a.(^-^).  (2) 

Short  Columns.      Where  ?i>  =:the  safe  total  load  in  pounds. 
Where  a  =  the  area  of  cross-section  in  inches. 
And  where  (  -^  j  ^  the  safe  resistance  to  crushing  per  square  inch. 

Example. 
What  is  the  safe  load  which  the  granite  cap  of  a  12"  x  12"  pier  tvill 
carry,  the  cap  being  twelve  inches  thick  ? 

The  cap  being  only  twelve  inches  high,  and  as  wide  and  broad  as 

1  This  is  not  theoretically  correct,  as  there  is  in  every  case  a  tendency  for  the 
material  under  compression  to  spread;  but  it  is  near  enough  for  all  practical 
purposes. 


COMPRESSION. 


23 


higli,  is  evidently  a  short  piece  under  compression,  therefore  the 
above  formiihi  (2)  applies. 

The  area  is,  of  course:  a=:  12.12=  144  square  inches. 

The  ultimate  resistance  of  granite  to  crushing  per  square  inch  is, 
say,  fifteen  thousand  pounds,  and  using  a  factor-of-safety  of  ten,  we 
have  for  the  safe  resistance :  — 

c        15000       i-ArtU 

—    = zrzloOOIbs. 

/  10 

Therefore  the  safe  load  w  on  the  block  would  be  :  — 

7«  =  144,  1500  =  216000  pounds. 

Where  long  pieces  (pillars)  are  under  compres-  Compression, 
1  ^  ,  •     i.     •  IT  1  Long  Columns. 

sion,  and  are  not  secured  against  yielding  sideways, 

it  is  evident  they  would  be  liable  to  bend  before  breaking.  To  ascer- 
tain the  exact  strain  in  sucli  pieces  is  probably  one  of  the  most  dif- 
ficult calculations  in  strains,  and  in  consequence  many  authors  have 
advanced  different  theories  and  formulae.  The  writer  has  always 
preferred  to  use  Rankine's  formula,  as  in  his  opinion  it  is  the  most 
reliable.  According  to  this,  the  greatest  strain  would  be  at  the  cen- 
tre of  the  length  of  the  pillar,  and  would  be  equal  to  the  load,  plus 
an  amount  equal  to  the  load  multiplied  by  the  square  of  the  length 
in  inches,  and  again  multij^lied  by  a  certain  constant,  ri,  the  whole 
divided  by  the  "  square  of  the  radius  of  gyration  "  of  the  cross-sec- 
tion of  the  pillar.  We  have  therefore  for  the  total  strain  at  the  cen- 
tre of  long  pillars  :  — 

I  w.l-n 
.  =  ..+  -^ 

Where   s  =  the  strain  in  pounds. 
"        ii)  =  the  total  load  in  pounds. 
"  Z  =  the  length  in  inches. 

<'        p-  =  the  s(piare  of  the  radius  of  gyration  of  the  cross-section. 
"        n  =  a  constant,  as  follows  :  — 
TABLE  II. 

VALUE  OF   n  IN  FORMULA  FOR    COMPRESSION. 


Material 

of 

pillar. 

Both  ends  of 

pillar  smooth 

(turned  or  planed.) 

One  end  smooth 
(turned  or  planed) 
other  end  a  pin  end. 

Both  ends 
pin  ends. 

Cast-iron 

0.C003 

0.0001 

0.00057 

■VVrought-iron 

0.000025 

0.000033 

0.00005 

Steel 

0.00002 

0.000025 

0.0000.33 

Wood 

0.00033 

0.00041 

0.00067 

Stone 

0.002 

Brick 

0.0033 

24  SAFE   BUILDING. 

The  stress  of  course  ■will  be  as  before  :  — 

V  =  a.  c. 
Where  r  =  the  ultimate  stress  in  pounds. 
"        a  =  the  area  of  cross-section  in  inches. 
"        c  =  the  ultimate  resistance  to  crushing  per  square  inch. 
Inserting  the  values  for  strain,  s,  and  stress,  v,  in  the  fundamental 
formula  (1)  we  have  :  — 

/     ,  w.r-.7i\ 


<j)=<^+'^) 


a 

w=  - 


(/)  (8) 


Long  Columns.        ^~| — 5^ 
P" 

Where  w  =  the  aafe  total  load  on  the  pillar. 
•'        a  =  the  area  of  cross-section  in  inches. 
"       p2  _-  tiie  square  of  the  radius  of  gyration  of  the  cross-section. 
"        /  =  the  length  in  inches. 
«  =  the  safe  resistance  to  crushing  per  square  inch. 

Example, 

What  safe  load  lo'dl  a  12"  x  12"  brick  pier  carry,  if  the  pier  is  ten 

feet  long,  and  of  good  masonry  f 

The  area  of  cross-section  will  be  :  — 

a  =  12.12  =  144  square  inches. 

The  square  of  the  radius  of  gyration  according  to  Section  No.  1 

in  Table  I  would  be :  — 

_,  and  as  d=l2,  we  have  n2—.ij:^±z  — 10 
12  12    ~ 

For  the  safe  resistance  to  crushing  per  square  inch,  we  have,  using 

a  factor-of- safety  of  ten,  and  considering  the  ultimate  resistance  to 

be  2,000  pounds  per  square  inch, 

_2000 

The  length  will  be  ten  feet,  or  one  hundred  and  twenty  inches ; 
therefore :  — 

Z2=  14400 
For  n  we  must  use  (according  to  Table  II),  for  brickwork :  — 
71  =  0.0033; 


/  c  \      2000      „^^ ,^ 
(7)= -10-=  200  lbs. 


COMPRESSION.  25 

Therefore  the  safe  total  load  on  the  pier  would  be :  — 

M^—.   14400.  0,0033        1+3,96 
"^  12 

In  all  formulae  where  constants  and  factors  of  safety  are  used,  it 
will  be  found  simpler  and  avoiding  confusion  to  immediately  reduce 
the  constant  by  dividing  it  by  the  factor-of-safety,  and  then  using 
only  the  reduced  or  safe  constant. 

Thus  if  c=:  48,000  pounds,  and  if /=4,  do  not  write  into  your 

formula  for  (-^)  =  iM2,  but  use  at  once  for  (-^)  =  12000. 

Materials  in  compression  that  have  an  even  bearing  on  all  parts  of 
the  bed  will  stand  very  much  more  compression  to  the  square  inch 
than  materials  with  rough,  uneven  or  rounded  beds,  or  where  the 
bearing  is  on  part  of  the  cross-section  only,  as  in  the  case  of  pins  (in 
trusses)  bearing  on  eye-bars.  It  is  usual  in  calculating  to  make 
allowance  for  this.  Columns  with  perfectly  even  bearing  on  all  parts 
of  the  bed  (planed  or  turned  iron  or  dressed  stone)  will  stand  the 
largest  amount  of  compression.  Columns  with  rough,  rounded  or 
uneven  ends  are  calculated  the  same  as  for  pin-ends  of  eye-bars.  In 
the  table  (II)  giving  the  values  for  n  of  Rankine's  formula  for  com- 
pression, the  different  values  for  smooth  and  also  for  pin  ends  are 
given. 

WBINKLIXG    STRAINS, 

Thin  pieces  of  wrought-iron  under  compression  endwise  may  neither 
crush  nor  deflect  (bend),  but  give  way  by  wrinkling,  that  is,  buckling 
or  corrugating,  provided  there  are  no  stiffening-rlbs  length- 
wise. 

Thus  a  square,  tubular   column,  if  the  sides  are  very' 
thin  might  give  way,  as  shown  in  Figure  2,  which  is  called 
wrinkling.     Or,  in  a  similar  way,  the  top  plate  of  a  boxed     Fig-  2. 
girder,  if  very  thin,  might  wrinkle,  as  shown  in  Figure  3,  under  heavy 
compressive  strains.     To  calculate  this  strain  use  the 
following  formula : 

Where  w  =  the  amount  of  ultimate  compression  in 
pounds  per  square  inch,  which  will  wrinkle  the  mate- 
rial. 

«>,.  =  a  constant, 

d    ■=  the  thickness  of  plate  in  inches, 


SAFE   BUILDING. 


b  =  the  unstiffened  breadth  of  plate  in  inches. 
If  a  plate  has  stiffening  ribs  along  both  edges,  use  for  b  the  actual 
breadth  between  the  stiffening  ribs ;  if  the  plate  is  stiffened  along 
one  edge  only,  use  4&,  in  place  of  6.  Thus,  in  the  case  of  the  boxed 
girder,  Figure  3,  if  we  were  considering  the  part  of  top  plate 
between  the  webs,  we  should  use  for  b  in  the  formula,  the  actual 
breadth  of  b  in  inches ;  while,  if  we  were  considering  the  overhang- 
ing part  &,  of  top  plate,  -we  should  use  4&,  in  place  of  b  in  formula. 
For  rectangular  columns  use  160,000  pounds  for  Wr',  for  tubular 
beams,  top  plates  of  girders,  and  single  plates  use  200,000  pounds  for 

Wr-     With  a  factor-of-safety  of  3,  we  should  have 


160000 


=  53000 


pounds  for  rectangular  columns,  and 


200000 


66000  pounds  for 


tubular  beams,  top  plates  of  riveted  girders  and  single  plates. 

o  p  f\r\f\ 

For  to  we  shall  use,  of  course, —  =  12000  pounds,  which  is 

the  safe  allowable  compressive  strain.  This  would  give  the  following 
table  for  safe  unstiffened  breadth  of  wrought-iron  jilates,  to  prevent 
wrinkling  of  plates. 

TABLE  III. 


Safe  breadth  in 

inches  of  Plate 

Safe  breadth 

n  inches  of  Plato 

stiffened  along  both  edges. 

stiffened  along  one  edge  only. 

(use  6.) 

(use  46,) 

Thickness  of 

Rectan- 

Tubular Beams, 

Rectangular 

Riveted  Girders 

Plate 

gular  Col- 

riveted Girders,  and 

Columns. 

and  single  Plates. 

In  inches. 

umns. 

single  Plates. 

1 

L                 -J 

K 

b 

TFT' 

V^  ^^ 

^ 

Tbil    t^I 

L 

h d      b i      r"        Ih 

IHj 

nljt        1^4 

lln^    "^-"H 

i 

2-"- 

^.3 
'^4 

8 

if 

X 

4' 

7  0 

\1 

1'' 

4 

^8 

'  1  6 

'4 

'■  8 

3 

7  5 

113 

U 

913 

'  16 

8, 

^ 

n 

151 

h\ 

h' 

1 

IVb 

18| 

3 

4H 

f 

14| 

22H 

m 

5| 

f 

1^6 

26i 

H 

6t^6 

1 

191 

301 

4| 

"fe 

H 

24f 

37if 

H 

9tV 

H 

29i 

45| 

h\ 

11 A 

If 

34^ 

52^1 

8t% 

13A 

2 

39 

60^ 

9f 

151 

The  above  table  will  cover  every  case  likely  to  arise  in  buildings. 


WRINKLING.  27 

Two  facts  should  be  noticed  in  connection  with  wrinkling: 

1.  That  the  length  of  plate  does  not  in  any  way  affect  the  resist- 
ance to  wrinkling,  which  is  dependent  only  on  the  breadth  and  thick- 
ness of  the  part  of  {jlate  unstiffened,  and 

2.  That  the  resistance  of  plates  to  wrinkling  being  dependent  on 
their  breadth  and  thickness  only,  to  obtain  equal  resistance  to  wrink- 
ling at  all  points  (in  rectangular  columns  with  uneven  sides),  the 
thickness  of  each  side  should  be  in  proportion  to  its  breadth. 

Thus,  if  we  have  a  rectangular  column  30"  X  15"  in  cross  section 
and  the  30"  side  is  1"  thick,  we  should  make  the  15"  side  but  ^"  thick, 
for  as  30"  :  1"  :  :  15"  :  ^". 

Of  course,  we  must  also  calculate  the  column  for  direct  crushing 
and  flexure,  and  in  the  case  of  beams  for  rupture  and  deflection,  as 
well  as  for  wrinkling. 

Example  of  Wrinkling. 

It  is  desired  to  make  the  top  plate  of  a  boxed  girder  as  wide  as  pos- 
sible, the  top  flange  is  to  be  1^"  thick,  and  is  to  be  subjected  to  the  full 
amount  of  the  safe  compressive  strain,  viz :  1 2,000  pounds  per  square 
inch  ;  how  wide  apart  should  the  ivebs  be  placed,  and  how  much  can  the 
plate  overhang  the  angles  without  danger  of  lonnkling  ?  Each  xceb  to  be 
I"  thick,  and  the  angles  4"  X  4"  each  ? 

For  the  distance  between  webs  we  use  b  in  Formula  (4). 

''— ^4-\^  12000/     —^4-02       — «i^6   » 
which  is  the  safe  width  between  webs  to  avoid  wrinklino- 

For  the  overhanging  part  of  top  plate  we  must  use  46,  in  place  of 
h  in  Formula  (4). 

^^•=  ^HtIU^J      =37if,  therefore, 

h.  =  ^    =  9,453,  or  say,     5.  =  9^^". 

The  total  width  of  top  plate  will  be,  therefore,  including  1"  for 
two  webs  and  8"  for  the  two  angles,  or  9",  and  remembering  that 
there  is  an  overhanging  part,  &„  each  side, 


am— CTa 


\^-m= 


G511". 

By  referring  to  Table  III,  we  should 
have  obtained  the  same  result,  with- 
F'£-  4  out  the  necessity  of  any  calculation. 

Figure  4  will  make  the  above  still  more  clear. 


28  SAFE   BUILDING. 

LATERAL     FLEXURE     IN     TOP    FLANGES    OF    BEAMS,    GIRDERS,    OR 
TRUSSES,   DUE    TO   COMPRESSION. 

The  usual  formulae  for  rupture  and  deflection  assume  tlie  beam, 
girder  or  truss  to  be  supported  against  possible  lateral  flexure 
(bending  sideways).  Now,  if  the  top  c-hord  of  a  truss  or  beam  is 
comparatively  narrow  and  not  supported  sideways,  the  heavy,  com- 
pressive strains  caused  in  same  may  bend  it  sideways.  To  calculate 
this  lateral  flexure,  use  the  formula  given  for  long  columns  in  com- 
pression, but  in  place  of  I  use  only  two-thirds  of  the  span  of  the 
beam,  girder  or  truss,  that  is  |/,  and  for  w  use  one-third  of  the  great- 
est compressive  strain  in  top  chord,  which  is  usually  at  the  centre. 

Inserting  this  in  Formula  (3)  we  have : 

"3"  ^TTA^  transposing,  we  have,  w  =—Wn     ^^^ 

where  a  the  area  of  the  cross-section  of  the  top  chord  in  inches, 

0^  is  the  square  of  the  radius  of  gyration  of  the  top  chord  around 
its  vertical  axis ;  we  must  therefore  reverse  the  usual  positions  of  h 
and  rf,  that  is  the  breadth  of  top  chord,  becomes  the  depth  or  d,  and 
the  depth  of  top  chord  becomes  the  thickness,  or  h  (both  in  formulae 
given  ia  last  column  of  Table  I.) 
—  is  the  greatest  allowable  compressive  strain  in   pounds  at  any 

point  to  resist  lateral  flexure  safel}'  at  that  point. 

(  —  )  is  the  safe  resistance  of  the  material   to    compression   per 

square  inch  in  pounds. 

I  is  the  total  length  of  span  in  inches. 

n  is  given  in  Table  II. 

Example. 

A  trussed  girder  is  60'  long  between  hearings,  and  not  supported  side- 
ways ;  the  top  chord  consists  of  two  plates  each  22"  deep  and  1"  thick; 
the  plates  are  2"  apart,  as  per  Figure  5.  The  greatest  compressive 
strain  on  top  chord  has  previously  been  ascertained  to  be  on  the  central 
panel,  and  to  he  525000  pounds.  Is  there  danger  of  the  girder  bending 
sideways? 

The  girder  is  safe  against  lateral  flexure  so  long  as  the  strain  at 

centre  does  not  exceed  -^  in  Formula  (5). 
Now,  the  area  «^  2.1.22  =44. 


LATERAL   FLEXURE.  29 

Usin"'  48000  pounds  per  square  inch  for  ultimate  resistance  to  com- 
pression of  wrought-iron,  and  a  factor-of-safety  of  4,  we  have 

48000 


(7) 


=  12000 


4 
The  length  is  GO',  or  720",  therefore 

Z2  =  518400. 
From  Table  II  we  have 
?i  =  0,000025. 
And  from    Table  I,   section  Number   16,  we  have 
Fig-  5.        for  the  above  cross-section, 

§   ~~  V2id-d,) 
As  we  are  considering  the  section  for  bending  sideways,  we  must, 
of  course,  take  the  neutral  axis  x---y  vertically,  therefore  d  becomes 
4"  and  d  becomes  2".     This  supposes  the  plates  to  be  stiffly  latticed 
or  bolted  together,  with  separators  between.     We  have  then 

43-23 

Then  for  w  we  have, 

3.44.12000 


«>=  4,518400  0.000025. 

9-     21- 

_  1584000     _  1584000     _  456  484  lbs. 
1+2,47  3,47 

Or,  we  find  that  there  is  danger  of  the  girder  bending  sideways 
long  before  the  actual  compressive  strain  of  525000  pounds  has  been 
reached.  It  will,  therefore,  be  necessary  to  re-design  the  top  chord, 
so  that  it  will  be  stiffer  sideways.  This  subject  will  be  more  fully 
treated  when  considering  trusses. 

TENSION. 

In  tension  the  load  is  applied  directly  to  the  material,  and  it  is, 
therefore,  evident  that  no  matter  of  what  shape  the  material  may  be, 
the  strain  will  always  be  the  same.     This  strain,  of  course,  will  be 
just  equal  to  the  load,  and  we  have,  therefore:  — 
s-=w. 

Where  s  =  the  amount  of  strain. 

Where  w  =  the  amount  of  load. 

The  weakest  point  of  the  piece  under  tension  will,  of  course,  be 
where  it  has  the  smallest  area  of   cross-section;   and  the  stress  at 


30  SAFE   BUILDING. 

such  point  will  be  equal  to  the  area  of  cross-section,  multiplied  by  the 
amount  of  resistance  its  fibres  are  capable  of.^ 

The  amount  of  resistance  to  tension  the  fibres  of  a  material  are 
capable  of  is  found  by  experiments  and  tests,  and  is  given  for  each 
material  per  square  inch  of  cross-section.  A  table  of  constants  for 
the  ultimate  and  safe  resistances  to  tension  of  different  materials 
will  be  given  later;  in  all  the  formulas  these  constants  are  represented 
by  the  letter  /. 

We  have,  then,  for  tlie  stress :  — 
v  =  a.  t 

Where  i;  =  the  amount  of  ultimate  stress. 

Where  a=:the  area  of  cross-section. 

Where  i=  the  ultimate  resistance  to  tension,  per  square  inch  of  the 
material. 

Therefore,  the  fundamental  formula  (1),  viz. :  v=rs.f,  becomes 
for  pieces  under  tension :  — 

a.  t  =  w.  f,  or :  — 

t«  =  a.(-i)  (6) 

Where  7i>  =  the  safe  load  or  amount  of  tension  the  piece  will  stand. 
Where  a  =  the  area  of  cross-section  at  the  weakest  point  (in  square 
inches). 

Where  (A'j^the  safe  resistance  to  tension  per  square  inch  of 

the  material. 

Example. 

A  weight  is  hung  al  the  lower  end  of  a  vertical  wrought-iron  rod, 
wMch  is  firmly  secured  at  the  other  end.  The  rod  is  3''  at  one  end  and 
tapers  to  2"  at  the  other  end.  How  much  weight  will  the  rod  safely 
carry  f 

The  smallest  cross-section  of  the  rod,  where  it  would  be  likely  to 
break,  would  be  somewhere  very  close  to  the  2"  end,  or,  say,  2"  in 
diameter.     Its  area  of  cross-section  at  this  point  will  be  :  — 

99  9*^  •      t 

a=  —         —     =  3i  square  inches. 
7*4  '     ^ 

The  ultimate  resistance  to  tension  of  wrought-iron  per  square  inch 

is,  from  forty-eight  thousand  pounds  to  sixty  thousand  pounds.    We 

do  not  know  the  exact  quality,  and,  therefore,  take  the  lower  figure  ; 

1  This,  again,  is  not  theoretically  correct,  as  a  piece  under  tension  is  apt  to 
stretcli  and  so  reduce  the  area  of  its  cross-section  ;  hut  the  above  is  sufiiciently 
correct  for  all  practical  purposes. 


SHEARING. 


31 


using  a  factor-of-safety  of  four,  we  have  for  the  safe  resistance  to 
tension  per  square  inch  :  — 

/J^X       48  000     _  12  000  pounds. 

Therefore,  the  safe  load  will  be  :  — 

w^3\.     12  000     z=  37  714  pounds. 
SHEARING. 

In  compression  the  fibres  are  shortened  by  squeezing;  in  tension 
they  are  elongated  by  pulling.  In  shearing, 
however,  the  fibres  are  not  disturbed  in  their 
individualities,  but  slide  past  each  other. 

When  this  sliding  takes  place  across  the  grain 
of  the  fibres,  the  action  of  shearing  is  more  like 
cutting  across.  When  this  sUding  takes  place 
Fig.  6.  '^'^'^  along  the  grain,  tlie  action  of  sliearing  is  more 
like  splitting.  Thus,  if  I  very  deep,  but  thin,  beam  is  of  short  span 
and  heavily  loaded,  it  might  not  break  transversely,  nor  deflect  ex- 
cessively, but  shear  off  at  the  supports,  as  shown  in  Figure  6,  the  ac- 
tion of  the  loads  and  supports  being  like  a  large  cutting-machine, 
the  weights  cutting  off  the  central  part  of  beam  and  forcing  it  down- 
wards past  the  support.     This  would  be  shearing  across  tlie  grain. 

If  the  foot  of  a  main  rafter  is  toed-in  to  the  end  of  a  tie-beam,  and 
the  foot  forces  its  way  outwardly,  pushing  away  the  block  or  part  of 
tie-beam  resisting  it  (splitting  it  out  as  it  were),  this  would  be  shear- 
ing along  the  grain. 

In  most  cases  (except  in  transverse  strains)  the  load  is  directly  ap- 
plied to  the  point  being  sheared  off ;  the  strain  will,  therefore,  just 
equal  the  load,  and  we  have  :  — 

Where  s^  the  amount  of  the  shearing  strain. 
"     if>  =       "  "       load. 

The  stress  will  be  equal  to  the  area  of  cross-section  (affected  by 
the  shearing  strain)  multiplied  by  the  amount  of  resistance  to  sep- 
aration from  each  other  that  its  fibres  are  capable  of. 

This  amount  of  resistance  is  found  by  tests  and  experiments,  and 
is  given  for  each  material  per  square  inch  of  cross-section.  A  table 
of  constants  for  resistance  to  shearing  of  different  materials  will  be 
given  later ;  in  the  formulae  these  constants  are  represented  by  the 
letter  g  for  shearing  across  the  grain,  and  g,  for  shearing  along  the 
grain. 


32  SAFE   BUILDING. 

We  have,  then,  for  the  stress  :  — 

v  =  a.  g. 
Where  v  =  the  amount  of  ultimate  stress. 
Where  a  =  the  area  of  cross-section  in  square  inches. 
Where  (7  =  the  ultimate  resistance  to  shearing  across  the  grain  per 
square  inch. 

Therefore,  the  fundamental  formula  (1)  v  =  s.f,  becomes  for  pieces 
under  shearing  strains  across  the  grain  :  — 
a.  g  =  w.f,  or:  — 

(7) 


'=«•(!) 


/- 
And  similarly,  of  course,  we  shall  find  :  — 


,=..(x.) 


//  (') 

Where  w  =  the  safe-load. 

Where  a  =  the  area  of  cross-section  in  square  inches,  at  the  point 
where  there  is  danger  of  shearing. 

Where  {^^  =  the  safe-resistance  to  shearing  across  the  fibres  per 

square  inch. 

Where  ^  -^  j  =  the  safe-resistance  to  shearing  along  the  fibres  per 

square  inch. 

Example. 

At  the  lower  end  of  a  vertical  wrouglit-iron  fiat  bar  is  suspended  a 
load  of  eight  thousand  pounds.  The  bar  is  in  two  lengths,  riveted  to- 
gether with  07ie  rivet.     What  diameter  should  the  rivet  be  f 

The  strain  on  the  rivet  will,  of  course,  be  a  shearing  strain  across 
the  grain,  and  will  be  equal  to  the  amount  of  tension  on  the  bar, 
which  we  know  is  equal  to  the  load.  We  use  Formula  (7),  and  have:  — 
w  =  8000  pounds. 

The  safe  shearing  for  wrought-iron  is  about  ten  thousand  pounds 
per  square  inch  ;  inserting  this  in  formula,  we  have  :  — 

8000  =  a.  10000,  or  a.  =  ^^      =f 

The  area  of  rivet  must,  therefore,  be  four-fifths  of  a  square  inch. 
To  obtain  diameter,  we  know  that :  — 

d=s/w^  =\/m  =^n  =^1^01818- 

This  is,  practically,  equal  to  one  ;  therefore,  the  diameter  of  rivet 
should  be  1". 


CROSS-SHEARING    IX   BEAMS.  83 

In  transverse  strains  the  (vertical)  cross-sliearing  is  generally  not 
equal  to  the  load,  but  varies  at  different  points  of  the  beam  or  canti- 
lever. The  manner  of  calculating  transverse  strains,  however,  allows 
for  straining  only  the  edges  (extreme  fibres)  up  to  the  maximum ;  so 
that  the  intermediate  fibres,  not  being  so  severely  tested,  generally 
have  a  sufiicient  margin  of  unstrained  strength  left  to  more  than  off- 
set the  shearing  strain.  In  solid  beams  it  can,  therefore,  as  a  rule, 
be  overlooked,  except  at  the  points  of  support.  (In  plate-girders  it 
must  be  calculated  at  the  different  jjoints  where  weights  are  applied.) 
The  amount  of  the  shearing  at  each  support  is  equal  to  the  amount 
of  load  coming  on  or  carried  by  the  support. 

"We  must,  therefore,  substitute  for  w  in  Formula  (7)  either />  or  q, 
as  the  case  may  be,  and  have  at  the  left-hand  end  of  beam  for  the 
safe  resistance  to  shearing  :  — 


.(5) 


And  at  the  ridit-haud  end  of  beam  :  — 


=  a.U) 


(9) 


(10) 


Where /)  =  the  amount  of  load,  in  pounds,  carried  on  the  left- 
hand  support. 

Where  q  =  the  amount  of  load,  in  pounds,  carried  on  the  right- 
hand  support. 

Where  a  =  the  area  of  cross-section,  in  inches,  at  the  respective 
support. 

Where  ^  -^  j  =  the  safe  resistance,  per  square  inch,  to  cross-shear- 
ing. 

Example. 

A  spruce  beam  of  5'  dear  span  is  24"  deep  and  3"  wide  ;  how  much 
uniform  load  will  it  carry  safely  to  avoid  the  danger  of  shearing  off  at 
either  point  of  support? 

The  beam  being  uniformly  loaded,  the  supports  will  each  carry 
one-half  of  the  load ;  if,  therefore,  we  find  the  safe  resistance  to  shear- 
ing at  either  support,  we  need  only  double  it  to  get  the  safe  load  (in- 
stead of  calculating  for  the  other  support,  too,  and  adding  the  results). 

Let  us  take  the  left-hand  support.   From  Formula  (9)  we  have  :  — 

Now,  we  know  that  a  ==  24.  3  =  72  square  inches. 


84  SAFE   BUILDING. 

The  ultimate  resistance  of  spruce  to  cross-shearing  is  about  thirty- 
six  hundred  pounds  per  square  inch  ;  using  a  factor-of-safety  of  ten, 
we  have  for  the  safe  resistance  per  square  inch:  — 

We  have,  now  :  — 

p  =  72.360  =  25920  pounds. 
Similarly,  we  should  have  found  for  the  right-hand  support :  — 
^  =  25920  pounds.     And  as:  — 
u  =p  -^q  =  51840  pounds, 
that  will,  of  course,  be  the  safe  uniform  load,  so  far  as  danger  of  shear- 
ing is  concerned. 

The  beam  must  also  be  calculated  for  transverse  strength,  deflec- 
tion and  lateral  flexure,  before  we  can  consider  it  entirely  safe. 
These  will  be  taken  up  later  on. 

Should  it  be  desired  to  find  the  amount  of  vertical  shearing  strain 
X  at  any  point  of  a  beam,  other  than  at  the  points  of  support,  use  :  — 

ovi—^w  (11) 

Where  a:  =  the  amount  of  vertical  shearing  strain,  in  pounds,  at 
any  point  of  a  beam. 

C  „  ■)        the  reaction,  in  pounds,  (that  is,  the  share  of  the 
Where  ^  or  >-  =  total  loads  carried)  at  the  nearer  support  to  the 

(  '7  )        point. 
Where  S  ?«  =  the  sum  of  all  loads,  in  pounds,  between  said  nearer 
support  and  the  point. 

When  X  is  found,  insert  it  in  place  of  w;,  in  Formula  (7),  in  order 
to  calculate  the  strength  of  beam  necessary  at  that  point  to  resist  the 

shearing. 

Exa?nple. 

A  spruce  beam,  20'  long,  and  8"  deep,  carries  a  uniform  load  of  one 
hundred  pounds  per  running  foot.  WJiat  should  be  the  thickness  of 
beam  b'  from  either  support,  to  resist  safely  vertical  shearing? 

Each  support  will  carry  one-half  the  total  load  ;  that  is,  one  thou- 
sand pounds;  so  that  we  have  for  Formula  (11)  :  — 

|or  >-"=:1000  pounds. 

The  sum  of  all  loads  between  the  nearer  support  and  a  point  5' 
from  support  will  be :  — 

S  ivz=b.  100  =  500  pounds. 


HORIZONTAL    SHEARING    IN    BEAMS.  35 

Therefore,  the  amount  of  slieariiig  at  the  point  5'  from  support  will 
be:  — 

a:  =:  1000  —  500  =  500  pounds. 
Inserting  this  in  Forrauhi  (7)  we  have  :  — 

500 
500  =  «.  (y),  or,  (1=  /  £\ 

We  have  just  found  that  for  spruce, 
fJl-.^  =  360  pounds. 

Therefore,   a  =  ' —    i=:  1,39  square  inches. 
3G0  '         ^ 

And,  as  b.  d  =  a,  ov  b  =  — ,    we  have,  b  =  -!— -  =  -— 
d  8  6 

This  is  such  a  small  amount  that  it  can  be  entirely  neglected  in  an 
8"  wooden  beam. 

To  find  the  amount  of  vertical  shearing  at  any  point  of  a  canti- 
lever, other  than  at  the  point  where  it  is  built  in,  use :  — 

x=  S  10  (12) 

Where  x  the  amount  of  vertical  shearing  strain,  in  pounds,  at  any 
point  of  canti-Iever. 

Where  S  lu  the  sum  of  all  loads  between  the  free  end  and  said 
point. 

To  find  the  strength  of  beam  at  said  point  necessary  to  resist  the 
shearing,  insert  x  for  w  in  Formula  (7). 

In  transverse  strains  there  is  also  a  horizontal  shearing  along  the 
entire  neutral  axis  of  the  piece.  Tliis  stands  to  reason,  as  the  fibres 
above  the  neutral  axis  are  in  compression,  while  those  below  are  in 
tension,  and,  of  course,  the  result  along  the  neutral  line  is  a  tendency 
of  the  fibres  just  above  and  just  below  it,  to  slide  past  each  other  or 
to  shear  off  along  the  grain. 

AVe  can  calculate  the  intensity  (not  amount)  of  this  horizontal  shear- 
ing at  any  point  of  the  piece  under  transverse  strain. 

If  X  represents  the  amount  of  vertical  shearing  at  the  point,  then 

3     X 

the  intensity  of  horizontal  shearing  at  the  point  is  = . 

A     a 

If  this  intensity  of  shearing  does  not  exceed  the  safe-constant  \f} 

for  shearing  along  the  fibres,  the  piece  is  safe,  or  :  — 


3 


-M^)  (»> 


2'     a  —\f 


36  SAFE    BL'ILDIXa. 

Where  X  is  found  by  formulic  (11)  or  (12)  for  any  point  of  beam 

or, 

,„,  \  ^'  ,'  the  amount  of  supnortino;  force,  in  pounds, 

W  here  x  —   -;  uv  ^   —  ^  ^  ^^ 

(  q  [  foi"  either  point  of  support. 

Where  a  =  the  urea  of  cross-section  in  square  inches. 

Where  (  2;  )  =  the  amount  of  safe  resistance,  per  sqnare  inch,  to 

shearing  along  fibres. 

Example, 
Take  the  same  beam  as  he/ore.     The  amount  of  vertical  shearing  5' 
from  support  ive found  to  bejice  laindred pounds,  or:  — 
X  =  500. 
The  area  was  8"  multiplied  by  thickness  of  beam,  or  :  — 

a  =  8b. 
The  ultimate  shearing  along  the  fibres  of  spruce  is  about  four  hun- 
dred pounds  per  square  inch,  and  with  a  factor-of-safety  of  ten,  we 
should  have :  — 

\f)         10 
Inserting  this  in  Formula  (13)    ^  .   —  =  40 

,        1500      0//Q1 

1G.40 
The  beam  should,  therefore,  be  at  least  2^"  thick,  to  avoid  danger 
of  longitudinal  shearing  at  this  point.  At  either  point  of  support 
the  vertical  shearing  will  be  equal  to  the  amount  supported  there ; 
that  is,  one-half  the  load,  or  one  thousand  pounds.  Substituting  this 
for  X  in  Formula  (13),  we  have:  — 

l.i^^40,ori=:.^^-^     =4,"68. 
2         8&  1G.40 

The  beam  would,  therefore,  have  to  be  4|"  thick  at  the  points  of 
support,  to  avoid  danger  of  longitudinal  shearing.  The  beam,  as  it  is, 
is  much  too  shallow  for  one  of  such  span,  a  fact  we  would  soon  dis- 
cover, if  calculating  the  transverse  strength  or  deflection  of  beam, 
which  will  be  taken  up  later  on.  It  will  also  be  found  that  the  greater 
the  depth  of  the  beam,  the  smaller  will  be  the  danger  from  longitu- 
dinal shearing,  and,  consequently,  to  use  thinner  beams,  it  would  be 
necessary  to  make  them  deeper. 


TABLK    IV. 


37 


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45 


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46 


SAFE    BUILDING. 


The  amounts  given  in  Table  IV  for  compression,  tension  and  shear- 
ing are  along  fibres,  except  where  marked  across. 

It  will  also  be  noticed  that  the  factors -of-sa£ety  chosen  are  very  dif- 
ferent; the  reason  being  that  where  figures  seemed  reliable  the  fac- 
tor chosen  was  low,  and  became  higher  in  proportion  to  the  unre- 
liability of  the  figures.  The  tables,  as  they  are,  are  extremely  un. 
satisfactory  and  unreliable,  though  the  writer  has  spent  much  time 
in  their  construction.  Any  one,  who  will  devote  to  the  subject  even 
the  slightest  research,  will  find  that  there  are  hardly  any  two  origi- 
nal experimenters  who  agree,  and  in  most  cases,  the  experiments  are 
so  carelessly  made  or  recorded  that  they  are  of  but  little  value. 


TABLE  VI. 

WEIGHT  PER  CUBIC  FOOT  OF  MA.TERIALS. 

(Xot  included  in  Tables  IV  and  V.) 


Material. 


Ashes 

Asphali 

Butter 

Camphor 

Charcoal 

Coal,  solid 

"      loose 

Coke 

Cork 

Cotton  in  bales 

Fat 

Gunpowder... . 
Hay  in  bales.. . 

Isinglass 

Lead,  red 

Paper 


Weight. 


59 
150 
60 
63 
23 
93 
64 
60 
15 
20 
58 
66 
17 
70 
660 
65 


Material. 


Peat 

Petritied  wood 

Pitch 

Plumbago 

Pumice-ttone 

Resin 

Rock  crystal 

Rubber 

Salt 

Saltpetre ' 

Snow,  fresh  fallen. 

"    solid 

Sugar 

Sulphur 

Tiles 

Water 


Weight. 


So 
145 

75 
131 

66 

68 
172 

62 

134 

130 

6 

20 

82 
125 
115 

63 


REAC'IION   OF   SUPPORTS. 


47 


TRANS VKRSK    STRKNGTII.  —  RL'PTURE. 

If  a  beam  is  supported  at  two  en<]s,  and  loads  are  applied  to  the 
beam,  it  is  evident :  — 

1st,  that  the  beam  will  bend  under  the  load,  or  deflect. 

2d,  that  if  the  loading  continues,  the  beam  will  eventually  break, 
or  be  ruptured. 

Deflection  when  "^'"^  methods  of  calculating  deflection  and  rup- 
non-important.  ture  diff'jr  very  greatly.  In  some  cases,  -where  de- 
flection in  a  beam  would  do  no  damage — such  as  cracking  plaster, 
lowering  a  column,  making  a  floor  too  uneven  for  machinery,  etc., — or 
where  it  would  not  look  unsightly,  we  can  leave  deflection  out  of  the 
question,  and  calculate  for  rupture  only.  Where,  however,  it  is  im- 
portant to  guard  against  deflection,  we  must  calculate  for  both. 

RKACTIOX   OF   SUPPORTS. 

If  we  imagine  the  loaded  beam  supported  at  both  ends  by  two 
giants,  it  is  evident  that  each  giant  would  have  to  exert  a  certain 
amount  of  force  upwards  to  keep  his  end  of  the  beam  from  tipping. 

We  can  therefore  imagine  in  all  cases  the  supports  to  be  res'isting 

Amount  of  Reac-  *^''  reacting  with  force  sufficient  to  uphold  their  re- 

*'0"'  spective   ends.     The  amount  of  this  reaction  for 

either  support  is  equal  to  the  load  multiplied  by  its  distance  from  the 

further  support,  the  whole  divided 
by  the  length,  or 

w.  n 


p: 


I 


(14) 


Where  /J  =  the  amount  of  the  left 
hand  reaction  or  supporting  force. 

and  q=z——.       (15) 

Where  q  =  the  amount  of  the  right 
hand  reaction  or  supporting  force. 
If  there  are  several  loads  the  same  law  holds  good  for  each,  the 
reaction  being  the  sum  of  the  products,  or 


?r,  n    I    20,,  s 


(16) 

(17) 


andq  =  -Y  +  -^-f- 

As  a  check  add  the  two  reactions  together  and  their  sum  must 
equal  the  whole  load,  that  is,  p-\-q  =  lo,  -|-  ii\, 


48 


SAFE  BUILDING. 


Example. 
A  beam  9'  2"  long  between 
bearings  carries  two  loads, 
one  of  200  lbs.  4'  2"  from 
the  left-hand  support,  and 
the  other  of  300  lbs.  3'  4" 
from  the  right-hand  support. 
What  are  the  right-hand^ 
arid  left-hand  reactions  ? 

Referring    to   Figure   8  Fig.  8. 

we  sliould  have  w,  =  200  lbs.,  and  ?r„  =  300  lbs.,  further  Z  =:  110"  ; 
m  =  50";  n  =  60";  s  =  40",  and  r=70",  therefore  the  left-hand 
reaction  would  be :  — 

200.  60    ,    300.  40 


p: 


=  2 18^2_  pounds. 


110       '       110 
and  the  right-hand  reaction  would  be  :  — 

200.  60    ,    300.  70        „oi  q  i. 

o  = + =  281  fir  pounds. 

^  110       '       110  11^ 

As  a  check  add  p  and  q  together,  and  they  should  equal  the  whole 

load  of  500  lbs.,  and  we  have  in  effect :  — 

p-\-  q  =  218^2_  _|_  281^9j-  =  500  pounds. 
If  the  load  on  a  beam  is  uniformly  distributed,  or  is  concentrated 
at  the  centre  of  the  beam,  or  is  concentrated  at  several  points  along 
the  beam,  each  half  of  beam  being  loaded  similarly,  then  each  sup- 
port will  react  just  one  half  of  the  total  load. 

THE   PRINCIPLE   OF   MOMENTS. 

Law  of  Lever.  The  law  of  the  lever  is  well  known.  The  distance 
of  a  force  from  its  fulcrum  or  point  where  it  takes  effect  is  called  its 
leverage.  The  effect  of  the  force  at  such  point  is  equal  to  the  amount 
of  the  force  multiplied  by  its  leverage. 

Moment  of  a  The  effect  of  a  force  (or  load)  at  any  point  of  a 
force.  beam  is  called  the  moment  of  the  force  (or  load)  at 

said  point,  and  is  equal  to  the  amount  of  the  force  (or  load)  multi- 
plied by  the  distance  of  the  force  (or  load)  from  said  point,  the 
distance  measured  at  right  angles  to  the  line  of  the  force.  If  therefore 
we  find  the  moments  —  for  all  of  the  forces  acting  on  a  beam  —  at 
any  single  point  of  the  beam  we  know  the  total  moment  at  said  point, 
and  this  is  called  the  bending-momenl  at  said  point.    Of  course,  forces 


BENDING    MOMENT. 


49 


actinr'  in  opposite  directions  will  give  opposite  mo- 
Bend  Ing  oii  »ri4i 

moment,  nients,  and  will  counteract  each  other;  to  hnd  tue 

bending-moment,  therefore,  for  any  single  i)oint  of  a  beam  take  the 

difference  between  the  sums  of  the  opposing  moments  of  all  forces 

acting  at  that  point  of  the  beam. 

Now  on  any  loaded  beam  we  have  two  kinds  of  forces,  the  loads 
which  are  pressing  downwards,  and  the  supports  which  are  resisting 
upwards  (theoretically  forcing  upwards).  Again,  if  we  imagine 
that  the  beam  will  break  at  any  certain  point,  and  imagine  one  side 
of  the  beam  to  bo  rigid,  while  the  other  side  is  ten  ling  to  break 
away  from  the  rigid  side,  it  is  evident  that  the  effect  at  the  point  of 
rupture  will  be  from  one  side  only ;  therefore  we  must  take  the  forces 
on  one  side  of  the  point  only.  It  will  be  found  in  practice  that  no 
matter  for  what  point  of  a  beam  the  bending  moment  is  sought,  the 
bending  moment  will  be  found  to  be  the  same,  whether  we  take  the 
forces  to  the  right  side  or  left  side  of  the  point.  This  gives  an  ex- 
cellent check  on  all  calculations,  as  we  can  calculate  the  bending 
moment  from  the  forces  on  each  side,  and  the  results  of  course  should 
be  the  same. 

Now  to  find  the  actual  strain  on  the  fibres  of  any  cross-section  of 
the  beam,  we  must  find  the  bending  moment  at  the  point  where  the 
cross-section  is  taken,  and  divide  it  by  the  moment  of  resistance  of 
the  fibre,  or, 

^=  s 
r 

Where  m  =  the  bending  moment  in  lbs.  inch. 

Where  r  =  the  moment  of  resistance  of  the  fibre  in  inches. 

Where  s  =  the  strain. 

The  stress,  of  course,  will  be  equal  to  the  resistance  to  cross-break- 
ing the  fibres  are  capable  of.  In  the  case  of  beams  which  are  of  uni- 
form cross-section  above  and  below  the  neutral  axis,  this  resistance  is 
called  the  Modulus  of  Rupture  (k).  It  is  found  by  experiments  and 
tests  for  each  material,  and  will  be  found  in  Tables  IV  and  V.  AVe 
have,  then,  for  uniform  cross-sections :  — 
V  =k 

Where  v  =  the  ultimate  stress  per  square  inch. 

Where  k  =■  the  modulus  of  rupture  per  square  inch.  Inserting 
this  and  the  above  in  the  fundamental  formula  (1),  viz.:  v^sf, 
we  have : — 

,  771/. 

«  =  —./,  or 


50  SAFE    BUILDING. 

Transverse  ^ 

strength      uni-  — -. —  =  r  (18) 

form  cross  sec-  { -L\ 

tion.  \fj 

Where  ?/i  =  the  bending  moment  in  lbs.  inch  at   a  given  point  of 
beam. 

Where  r  =  the  moment  of  resistance  in  inches  of  the  fibres  at  said 
23oint. 

Where  (  —.)  =the  safe  modulus  of  rupture  of  the  material,  per 

square  inch. 

If  the  cross-section  is  not  uniform  above  and  be- 
"'"sfre ngth ^  s  e  c-  lo^^  the  neutral  axis,  we  must  make  two  distinct 
tion  not  uni-  calculations,  one  for  the  fibres  above  the  neutral 
axis,  the  other  for  the  fibres  below ;  in  the  former 
case  the  fibres  would  be  under  compression,  in  the  latter  under 
tension.  Therefore,  for  the  fibres  above  the  neutral  axis,  the  ultimate 
stress  would  be  equal  to  the  ultimate  resistance  of  the  fibres  to  com- 
pression, or  v  =  c. 

Inserting  this  in  the  fundamental  formula  (1),  we  have:  — 


?« 


//6  y. 


Upper  fibres.  { —  \ 


(19) 


Where  ??i  =  the  bending  moment  in  lbs.  inch,  at  a  given  point  of 

beam. 
Where  r  =  the  moment  of  resistance  in  inches  of  the  fibres  at 

said  point. 

AVhere  (  —  j  =  the  safe  resistance  to  crushing  of   the  material, 

per  square  inch. 

For  the  Jihres  below  the  neutral  axis,  the  ultimate  stress  would  be 
equal  to  the  ultimate  resistance  of  the  fibres  to  tension,  or,  v=:  t. 

Inserting  this  in  the  fundamental  formula  (1)  we  have :  — 

t  =  — ./,  or 

-^  =  r  (20) 

Lower  fibres.  /  f  \  '■     -' 


(7) 


Where  m  =  the  bending  moment  in  lbs.  inch  at  a  given  point  of 
beam. 


GREATKST    liK.NDlNG    MOMKNT.  51 

Where  r  =  the  moment  of  resistance  in  inches  of  the   (il)res  at 
said  point. 

AYliere  (— )  =  the  safe  resistance  to  tension  of  the  material,  per 

square  inch. 

The  same  formulaj  apply  to  cantilevers  as  well  as  beams. 
Tlie  moment  of  resistance  r  of  any  fibre  is  equal  to  the  moment  of 
inertia  of  the  whole  cross-section,  divided  by  the  distance  of  the  fibre 
from  the  neutral  axis  of  the  cross-section. 

The  greatest  strains  are  along  the  upper  and 
Greatest  strains  lower  edges  of  the  beam  (the  extreme  fibres)  ;  we, 
on  extreme  fi-  therefore,  only  need  to  calculate  their  resistances, 
as  all  the  intermediate  fibres  are  nearer  to  the 
neutral  axis,  and,  consequently,  less  strained.  The  distance  of  fibres 
chosen  iu  calculating  the  moment  of  resistance  is,  therefore,  the  dis- 
tance from  the  neutral  axis  of  either  the  upper  or  lower  edges,  as  the 
case  may  be.  The  moments  of  resistance  given  in  the  fourth  column, 
of  Table  I,  are  for  the  upper  and  lower  edges  (the  extreme  fibres), 
and  should  be  inserted  in  place  of  r,  in  all  the  above  formulae. 

To  find  at  what  point  of  a  beam  the  greatest  bend- 
Point  of   great-  i"?  moment  takes  place  (and,  consequently,  the 
est        bending  greatest  fibre  strains,  also),  begin  at  either  support 
moment.  °^^  ^^^^  ^^^^^^  ^^^  ^^^^  towards  the  other  sup- 

port, passing  by  load  after  load,  until  the  amount  of  loads  that  have 
been  passed  is'cqual  to  the  amount  of  the  reaction  of  the  support 
(point  of  start)  ;  the  point  of  the  beam  wliere  this  amount  is  reached 
is  the  point  of  greatest  bending  moment. 

In  cantilevers  (beams  built  in  solidly  at  one  end  and  free  at  the 
other  end),  the  point  of  greatest  bending  moment  is  always  at  the 
point  of  the  support  (where  the  beam  is  built  in). 

In  light  beams  and  short  spans  the  weight  of  the  beam  itself  can  be 
neglect'^d,  but  in  heavy  or  long  beams  the  weight  of  the  beam  should 
be  considered  as  an  independent  uniform  load. 

RULES   FOR   CALCULATING   TRAXSVERSE   STRAINS. 

1.  Find  Reaction  of  each  Support. 
Summary  of  If  the  loads  on  a  girder  are  uniformly  or  sym- 

'*"'®®'  metrically  distributed,  each  support  carries  or  re- 
acts with  a  force  eciual  to  one-half  of  the  total  load.  If  the  weights 
are  unevenly  distributed,  each  support  carries,  or  the  reaction  of  each 
support  is  equal  to,  the  sum  of   the  products  of  each  load  into  its 


62  SAFE    BUILDING. 

distance  from  the  other  support,  divided  by  the  whole  length  of  span. 
See  Formula;  (14),  (15),  (16),  and  (17). 

2.  Find  Point  of  Greatest  Benxling  Moment. 

The  greatest  bending  moment  of  a  uniformly  or  symmetrically 
distributed  load  is  always  at  the  centre.  To  find  the.  point  of  great- 
est bending  moment,  when  the  loads  are  unevenly  distributed,  begin 
at  either  support  and  pass  over  load  after  load  until  an  amount  of 
loads  has  been  passed  equal  to  the  amount  of  reaction  at  the  support 
from  which  the  start  was  made,  and  this  is  the  desired  point.  In  a 
cantilever  the  point  of  greatest  bending  moment  is  always  at  the  wall. 
3.  Find  the  Amount  of  the  Greatest  Bending  Moment. 

In  a  beam  (supported  at  both  ends)  the  greatest  bending  moment 
is  at  the  centre  of  the  beam,  provided  the  load  is  uniform,  and  this 
moment  is  ecjual  to  the  product  of  the  whole  load  into  one-eighth  of 
the  length  of  span,  or 

rn=^^  ,  (21) 

Where  m  =  the  greatest  bending  moment  (at  centre),  in  lbs.  inch, 
of  a  uniformly-loaded  beam  supported  at  both  ends. 

Where  u  =  the  total  amount  of  uniform  load  in  pounds. 

Where  I  =  the  length  of  span  in  inches. 

If  the  above  beam  carried  a  central  load,  in  place  of  a  uniform 
load,  the  greatest  bending  moment  would  still  be  at  the  centre,  but 
would  be  equal  to  the  product  of  the  load  into  one-quarter  of  the 
length  of  span,  or 

m  =  —  (22) 

Where  m  =  the  greatest  bending  moment  (atxientre),  in  lbs.  inch, 
of  a  beam  with  concentrated  load  at  centre,  and  supported  at  both  ends. 

Where  w  =  the  amount  of  load  in  pounds. 

Where  I  =  the  length  of  span  in  inches. 

To  find  the  greatest  bending  moment  of  a  beam,  supported  at  both 
ends,  with  loads  unevenly  distributed,  imagine  the  girder  cut  at  the 
point  (previously  found)  where  the  greatest  bending  moment  is  known 
to  exist ;  then  the  amount  of  the  bending  moment  at  that  point  will 
be  equal  to  the  product  of  the  reaction  (of  either  support)  into  its 
distance  from  said  point,  less  the  sum  of  the  products  of  all  the  loads 
on  the  same  side  into  their  respective  distances  from  said  point,  i.  e., 
the  point  where  the  beam  is  supposed  to  be  cut.  To  check  the  whole 
calculation,  try  the  reaction  and  loads  of  the  discarded  side  of  the 
beam,  and  the  same  result  should  be  obtained. 


RULKS     TRANSVERSE   STRAINS. 


53 


Amount  of 
greatest  bend- 
ing moment. 


To  put  tlie  above  in  a  formula,  v;e  should  have :  — 

7)},,=p.x-'Z(u\  X,  -f  ?r„  x,,  -)-  «,•„,  a:,,,,  etc.)    (23) 
And  as  a  checL  to  above  : 

?«,v  =  q.(l-x)-^  ((i'„„  a-.„,  +  w,  X,  +  10^,  x,,) 
etc.  (24) 

Where  A  =  is    the  point  of  greatest  bending  moment. 
Where  m^^  is  the  amount  of  bending  moment,  in  lbs.  inches,  at  A. 
Where  ;j  =  is  the  left-hand  reaction,  in  pounds. 

Where  7  ^  is  the 
right-liand  reaction,  in 
pounds. 

Where  x  and  (/-.r) 
=  tlie  respective  dis- 
tances in  inches,  of 
the  left  and  right  reac- 
tions from  A. 

®  Where  x,,  a:,,,  a;,,,, 
etc.,  =  the  respective 
distances,    in     inches, 


r-Xi-  — [ — Xv-  -^ 


Fig.  9. 
from  A,  of  the  loads  to,,  w„,  w,„,  etc. 

Where  u\,  w,,,  ii\,„  etc.,  =  the  loads,  in  pounds. 

Where  S  z=  the  sign  of  summation. 

Tlie  same  formulfe,  of  course,  would  hold  good  for  any  point  of  beam. 

In  a  cantilever  (supported  and  built  in  at  one  end  only),  the  great- 
est bending  moment  is  alwai/s  at  the  point  of  support. 

For  a  uniform  load,  it  is  equal  to  the  product  of  the  whole  load 

into  one-half  of  the  length  of  the  free  end  of  cantilever,  or 

u.l  ,     ^ 

'"==-2-  (25) 

Where  m  z=  the  amount,  in  lbs.  inch,  of  the  greatest  bending 
moment  (at  point  of  support). 

Where  u  =z  the  amount  of  the  whole  uniform  load,  in  pounds. 

Where  I  =z  the  length,  in  inches,  of  the  free  end  of  cantilever. 

For  a  load  concentrated  at  the  free  end  of  a  cantilever,  the  great- 
est bending  moment  is  at  the  point  of  support,  and  is  equal  to 
the  product  of  the  load  into  the  length  of  the  free  end  of  canti- 
lever, or 

m  =  w.  I  (2G) 

Where  m  =  the  amount,  in  lbs.  inch,  of  the  greatest  bending 
moment  (at  j)oint  of  support). 


54  SAFE    BL'ILDIXG. 

Where  26"  :^  the  load,  in  pounds,  concentrated  at  free  end. 

Where  I  =  the  length,  in  inches,  of  free  end  of  cantilever. 

For  a  load  concentrated  at  any  point  of  a  cantilever,  the  greatest 
bending  moment  is  at  the  point  of  support,  and  is  equal  to  the  pro- 
duct of  the  load  into  its  distance  from  the  point  of  support,  or 

m  =  w.  X  (27) 

Where  m  =  the  amount,  in  lbs.  inch,  of  the  greatest  bending 
moment  (at  point  of  support). 

Where  w  z=  the  load,  in  pounds,  at  any  point. 

Where  x  ■=:■  the  distance,  in  inches,  from  load  to  point  of  support 
of  cantilever. 

Note,  that  in  all  cases,  when  measuring  the  distance  of  a  load,  we 
must  take  the  shortest  distance  (at  right  angles)  of  the  vertical 
neutral  axis  of  the  load,  (that  is,  of  a  vertical  line  through  the  centre 
of  gravity  of  the  load.) 

4.   Find  the  Required  Cross-section. 

To  do  this  it  is  necessary  first  to  find  what  wiU  be  the  required 
moment  of  resistance. 

If  the  cross-section  of  the  beam  is  uniform  above  and  below  the 
neutral  axis,  we  use  Formula  (18),  viz. :  — 


(I) 


k_ 

/' 

If  the  cross-section  is  unsymmetrical,  tliat  is,  not  uniform  above 
and  below  the  neutral  axis,  we  use  iov  i\\Q  fibres  above  the  ntutral  axis, 
formula  (19),  viz.:  — 


and  for  the  fibres  below  the  neutral  axis,  Formula  (20),  viz. 
m 


(f) 


In  the  latter  two  cases,  for  economj',  the  cross-section  should  be  so 
designed  that  the  respective  distances  of  the  upper  and  lower  edges 
(extreme  fibres),  from  the  neutral  axis,  should  be  proportioned  to 
their  respective  stresses  or  capacities  to  resist  compression  and  ten- 
sion.    This  will  be  more  fully  explained  under  cast-iron  lintels. 

A  simple  example  will  more  fully  explain  all  of  the  above  rules. 


EXAMPLE    TRANSVERSE    STRAINS 


55 


Example. 
Three  weights  of  respectivehj  500  lbs.,  1000  lbs.,  and  1500  lbs.,  are 
placed  on  a  beam  oflV  6"  (^or  210")  clear  span,  2'  6"  {or  30"),  V  6" 
{or  90"),  and  10'  0"  (or  120")  from  the  left-hand  support.  The  mod- 
ulus of  rupture  of  the  material  is  2800  lbs.  per  square  inch.  The  fuc- 
tor-of  safety  to  be  used  is  4.  The  beam  to  be  of  uniform  cross-section. 
What  size  of  beam  should  be  used? 

1.   Find  Reactions  {see  formulce  16  and  17). 

500.180   ,    1000.120   ,    1500.90 


Reaction  p  will  be  in  pounds,  = 
1642f  pounds. 


210 


Reaction  q  will  be  in  pounds,  =  — irV"  H~ 


210 
1000.90 


+ 


210 
1500.120 


210      '        210        '         210 
1357^  pounds. 

Check, p -\- q  must  equal  whole  load,  and  we  have  in  effect:  — 
p-\-q=lG42'f-}- 13571  =  3000,  which 
being  equal  to  the  sum  of  the  loads  is  correct,  for  :  — 
500-1-  1 000 -f- 1500  — 3000. 
2.    Find  Point  of  Greatest  Bending  Moment. 
Begin  at  jo,  pass  over  load  500,  plus  load  1000,  and  we  still  need 

to  pass  142|^  pounds  of 
^500)  2q^j  j.^  make  up  amount 

of  reaction  p  (1642| 
lbs.);  therefore,  the 
greatest  bending  mo- 
ment must  be  at  load 
1500;  check,  he^xn  at  9 
and  we  arrive  only  at 
the  first  load  (1500)  be- 
fore passing  amount  of 
reaction  q  (1357J  lbs.), 


(-3o^- 


■l8o4  -  ■- 


— -y^ 
-  -=i 


9^ 

•-\{2o- -4-- 9^- 

210'- 


Fig.   10. 


therefore,  at  load  1500  is  tiie  point  sought. 

3.  Find  Amount  of  Greatest  Bending  Moment. 
Suppose  the  beam  cut  at  load  1500,  then  take  the  left-hand  side  of 
beam,  and  we  have  for  the  bending  moment  at  the  point  where  the 
beam  is  cut. 

m  =  1642f  120—  (500.90  +  1000.30  -\-  1500.0)' 
=  197143— (45000-j-30000-f0) 
=  197143  —  75000 
==  122143  lbs.  inch. 


56  SAFE    BUILDING. 

As  a  check  on  the  calculation,  take  the  right-hand  side  of  beam  and 
we  should  have  :  — 

m  =  13571-.90  — 1500.0 
=  122143  —  0 
=  122143  lbs.  inch, 
which,  of  course,  proves  the  correctness  of  former  calculation. 
4.   Find  the  Required  Cross-section  of  Beam. 
We  must  first  find  the  required  moment  of  resistance,  and  as  the 
cross-section  is  to  be  uniform,  we  use  formula  (18),  viz. :  — 

m 
r 


i-f) 


Now,  m  =  122143,  and  -^  = =  700,  therefore, 

/  4 

r  =    "  ^  =  174,49  or  say  =  174,5 

700  J  J  y 

Consuming  Table  I,  fourth  column,  for  section  No.  2,  we  find 

r  =  —-,  we  have,  therefore, 
6 

—  =  174,5  or  5^2  _  1047. 
6 
If  the  size  of  either  6  or  rf  is  fixed  by  local  conditions,  we  can,  of 
course,  find  the  other  size  (d  or  h)  very  simply ;  for  instance,  if  for 
certain  reasons  of  design  we  did  not  want  the  beam  to  be  more  than 
4"  wide,  we  should  have 

6  =  4,  therefore,  4.^/2=1047,  and 

^2_  1047  _  .^^2^  therefore,  tZ=  (about)  16"', 

or,  if  we  did  not  want  the  beam  to  be  over  12"  deep,  we  should  have 
d  =  12,  and  d'^  =  12.12  =  144,  therefore, 

"&.144  =  104  7,  and  b  =  i^  =  7,2"  or  say  1\''. 
144 

The      deepest  One  thing   is  verv  important  and,  must  be  remem- 
beam   the   most  ,        ,     ,  ,        ,  '  ,       ,  , 

economical.  oered,  that  tiie  deeper  the  beam  is,  the  more  eco- 
nomical, and  the  stiffer  will  it  be.  If  the  beam  is  too  shallow,  it 
might  deflect  so  as  to  be  utterly  unserviceable,  besides  using  very 
much  more  material.  As  a  rule,  it  will  therefore  be  necessary  to 
calculate  the  beam  for  deflection  as  well  as  for  its  transverse  strength. 
The  deflection  should  not  exceed  0,"03  that  is. 
Safe  deflection.  ^^^reQ  one-hundredths  of  an  inch  for  ea(;h  foot  of 
span,  or  else  the  plastering  would  be  apt  to  crack,  we  have  then  the 
formula :  — 

8  =  1.  0,03  (28) 


DEFLECTION.  67 

Where  6  =  the  greatest  allowable  total  deflection,  in  inches,  at 
centre  of  beam,  to  prevent  [)laster  cracking. 

Where  L  =.  the  length  of  span,  in  feet. 

In  case  the  beam  is  so  unevenly  loaded  that  the  greatest  deflection 
will  not  be  at  the  centre,  but  at  some  other  point,  use :  — 

8  =-  A'.  0,06  (29) 

Where  6  =  the  greatest  allowable  total  deflection,  in  inches,  at 
point  of  greatest  deflection. 

Where  X  =  the  distance,  in  feet,  to  nearer  support  from  point  of 
greatest  deflection. 

If  the  beam  is  not  stiffened  sideways,  it  should  also  be  calculated 
for  lateral  flexure.  These  matters  will  be  more  fully  explained  when 
treating  of  beams  and  girders. 

COMPARATIVE     STRENGTH     AND     STIFFNESS     OF     BEAMS     AND 
COLUMNS. 

(1)  If  abeam  supported  at  both  ends  and  loaded  uniformly  will 
safely  carry  an  amount  of  load  =u;  then  will  the  same  beam  : 

(2)  if  both  ends  are  built  in  solidly  and  load  uniformly  distributed, 
carry  1^.  u, 

(3)  if  one  end  is  supported  and  other  built  in  solidly  and  load  uni- 
formly distributed,  cai-ry  1.  u, 

(4)  if  both  ends  are  built  in  solidly  and  load  applied  in  centre, 
carry  1.  w, 

(5)  if  one  end  is  supported  and  other  built  in  solidly  and  load  ap- 
plied in  centre,  carry  §.  u, 

(G)  if  both  ends  are  supported  and  load  applied  in  centre,  carry  ^.  u, 

(7)  if  one  end  is  built  in  solidly  and  other  end  free  (cantilever)  and 
load  uniformly  distributed,  carry  ^.  u, 

(8)  if  one  end  is  built  in  solidly  and  other  end  free  (cantilever)  and 
load  applied  at  free  end,  carry  ^.  it. 

That  is,  in  cases  (1),  (3)  and  (4)  the  effect  would  be  the  same  with 
the  same  amount  of  load ;  in  case  (2)  the  beam  could  safely  carry  1^ 
times  as  much  load  as  in  case  (1)  ;  in  case  (5)  the  beam  could  safely 
carry  only  §  as  much  as  in  case  (1),  etc.,  provided  that  the  length  of  span 
is  the  same  in  each  case.^ 

If  the  amount  of  deflection  in  case  (1)  were  <5,  then  would  the  amount 
of  deflection  in  the  other  cases  be  as  follows : 
Case  (2)  (5„  =  i.  (5,       Case  (4)  d,y  =  |.  <J,       Case  (7)  (5^„  =  9-?.  6, 
Case  (3)  <5„,  =  |.  d,        Case  (5)  6y  =  f .  d,        Case  (8)  6,,,,  =  25|.  d, 
Case  (6)(Sv.  =  lf'5, 

_  2  To  count  on  the  end  of  a  beam  being  built  in  solidly  would  be  very  bad  prac- 
tice in  most  cases  of  building  construction;  as,  for  instance,  a  wooden  beam  with 
end  built  in  solidly  cuki  no":  fall  out  in  caseof  fire, and  would  beapt  tothrowthe 
wall.  Even  where  practicable,  it  would  require  very  careful  supervision  to  get 
the  beam  built  in  properly  ;  then,  too,  it  causes  upward  strains  which  must  be 
overcome,  complicating  the  calculations  unnecessarily.  In  most  cases  where  it 
is  necessary  to  "build  in"  beam  ends,  the  additional  strength  and  diminished 
deflection  thereby  secured  had  better  be  credited  as  an  additional  margin  of 
safety.  The  above  rules  for  deflection  do  not  hold  good  if  the  beam  is  not  of 
uniform  cro«s-section  throughout ;  the  deflection  being  greater  as  the  variatiOD 
in  cross-section  is  greater. 


68  SAFE    BUILDING. 

TABLE 

BENDIXG-MOMENT   (Wl)   AND  AMOUNT  OF  SHEARING-STRAIN  (s) 


A 

^ L 


-<$ 


X 


U<?) 


^^ 


Load  at  any  point  of  Cantilever. 


Load  at  free  end  of 
cantilever. 


If  2/  ^^^       / 

smaller  thanl-:     greater  than—; 


Uniform  load  on 
cantilever. 


If  X  smaller  If  x,  greater    If  a;  =  ^ 
tlian  2/  use:    than  y  use:        use: 


For  X  use:     At  free 
end  use: 


For  X  use: 


At  free 
end  use: 


V- 

p 

> 

P 

Vi 

» 

II 

«r+ 

II 

CI- 

11 

w 

s 

o 

S 

o 

o 

09 

OO 

OO 

OO 

II 

» 

II 

II 

93 

II 

85 

WJH- 

S 

w|  H- 

& 

w|  i— 

Qc]  1— 

^ 

a 
a 

o 

a    S 

(I 
o 

a 

a 

o 

a 

'B 

p. 

a 

u. 

TABLE   VII. 


59 


VII.     (See  foot-uote  p.  CO.) 

OF  BEAMS  AND  CANTILEVEKS  FOR  VARIOUS  LOADS. 


-^ 


.-o* — HJ 


-0 


— ^TrVfl? 


F^ 


Load  at  any  point  on  beam  sup- 
ported at  both  ends. 


Load  at  centre  of 
beam  support- 
ed at  both  ends. 


Uniform  load  on 
beam  supported 
at  both  ends. 


Descrip- 
tion. 


If  M  smaller        If  w  greater 


than 


than  ±- 
2 


For  Xi  use:  For  x  use: 


At  load 


When  X  greater  than   l   use  (l-x) 
in  place  of  x.         "2" 


to)  fi  'O]   C; 


?        to|e 


5§ 


in 

II 

HI 

r»      '^ 

CD 

c 
0 

II         g"S|^ 

c^is    ^   lis 

•-J 

II 

t*|S 

0  p 

II 

t*I  8 

at  support 
p  or  (f. 

Location 
and  amount 
of  greatest 
shearing- 
strain  a. 

00 

00 

cy? 

1 

II 

II 

i-h  f 
0 

«c 

SS 

§ 

>l^l  ^ 

?= 

Oil 

» 

1?    g 

5*^ 

«« 

ex   c 

n, 

» 

a 

*■ 

a 

S-  0 

J^.i  « 

>-« 

«.|S 

D 

. 

a 

a    - 

=!■ 

0 

a 

a 

'^'l  'es 

a> 

«   c 

01 

t 

« 

60  SAFE    BUILDING. 

Strength  of  The  comparative  transverse  strength  of  two  or 
ferent  cross-sec-  ^^ove  rectangular  beams  or  cantilevers  is  directly 
*'°"®'  as  the  product  of  their  breadth  into  the  square  of 

their  depth,  provided  the  span,  material  and  manner  of  supporting 
and  loading  are  the  same,  or 

xz=bd^  (30) 

Where  a;  =  a  figure  for  comparing  strength  of   beams  of  equal 

spans. 
AVhere  h  =  the  breadth  of  beam,  in  inches. 
Where  d=  the  depth  of  beam,  in  inches. 

Example. 
What  is  the.  comparative  strength  between  a  3"  x  12"  beam,  and  a 
6"  X  12"  beam?   Also,  between  a  4"  x  12"  beam,  and  a  3"  x  16" beam? 
All  beams  of  same  material  and  span,  and  similarly  supported  and 
loaded. 

The  strength  of  the  3"  x  12"  beam  would  be 

ar,=:3.  12.  12  =  432. 
The  strength  of  the  6"  x  1 2"  beam  would  be 
a:„  =  6.  12.  12  =  864,  therefore, 
the  latter  beam  would  be  just  twice  as  strong  as  the  former. 
Again,  the  strength  of  the  4"  x  12"  beam  would  be 
x„.  =  4.  12.  12r=576 
and  the  strength  of  the  3"  x  16"  beam  would  be 
a;.,„  =  3.  16.  16=768. 

The  latter  beam  would  therefore  be or  iust  li  times  as  strono'  as 

576       •■  ^  ° 

the  former,  while  the  amount  of  material  in  each  beam  is  the  same,  as 

4.  12  =  3.  16  =  48  square  inches  in  each. 

The  reason  the  last  beam  is  so  much  stronger  is  on  account  of  its 

greater  depth. 

Strength  of  The  comparative  transverse  strength  of  two  ur 
beams  of  dif-  ,  .-i  c  •  , 

ferent  lengths,     niore  heams  or  cantilevers  of  same  cross-section  and 

material,  but  of  unequal  spans,  is  inversely  as  their  lengths,  provided 

manner  of  supporting  and  loading  are  the  same.     That  is,  a  beam  of 

twenty-foot  span  is  only  half  as  strong  as  a  beam  of  ten-foot  span,  a 

quarter  as  strong  as  one  of  five-foot  span,  etc. 

AH  measurements  in  Table  VII  are  in  inches;  all  weights  in  pounds;  c  =  mod- 
ulus of  elasticity  in  pounds  inch  ;  i=:  moment  of  inertia  of  cross-section  of  beam 
or  cantilever  around  its  neutral  axis  in  inches  ;  ??i  =  banding-moment  in  pounds 
Inch;  s=  amount  of  shearing  strain  in  pounds;  6  =  total  amount  of  deflection  in 
Inches. 


COMPAKATIVK    STIFFNESS.  61 

Stiffness  Of  The  stiffness  of  beams  or  cantilevers  of  same 
ferent  lengths.'  fross-section  and  material  (and  similarly  loaded  and 
supported),  however,  diminishes  very  rapidly,  as  the  length  of  span 
increases,  or  what  is  the  same  thing,  the  dellection  increases  much 
more  rapidly  in  proportion  than  the  length;  the  comparative  stiff- 
ness or  dellection  being  directly  as  the  cube  of  their  respective 
lengths  or  Z*. 

That  is  if  a  beam  10  feet  long  deflects  under  a  certain  load  one- 
third  of  an  inch,  the  same  beam  with  same  load,  but  20  feet  long  will 
deflect  an  amount  x  as  follows : 

a; :  i  =  208: 103,  or  a:  =  ?^  =  ^^=  2r 
^  1U3         3000  ^ 

Stiffness  of 
beams  of  dif-       J  he  comparative  stiffness,  that  is  amount  of  de- 
ferent cross-sec-  n     t-         c  .  i,  -i  •    • 

tions.  uection  ot  two  or  more  beams  or  cantilevers,  simi- 

larly supported  and  loaded,  and  of  same  material  and  span,  but  of 
different  cross-sections,  is  inversely  as  the  product  of  their  respective 
breadths  into  the  cubes  of  their  respective  depths  or 

Where  x  =  a  figure  for  comparing  the  deflection  of  beams  of  same 
material,  span  and  load. 

Where  6=^  the  breadth  of  beam,  in  inches. 
Where  c?=  the  depth  of  beam,  in  inches. 

Example. 

If  a  beam  3"  x  8"  deflects  ^"  under  a  certain  load,  ivhat  will  a  beam 
4"  X  12"  dcjlect,  if  of  same  material  and  span,  similarly  supported  and 
with  same  load  f 

For  the  first  beam  we  should  have 

For  the  second  beam  we  should  have 

X..  =  -i—  =     ^     =  0,00014 
"       4.123       6912         ' 

The  deflection  of  the  latter  beam  will  be  as 

S  :  0",5  =  0,00014  :  0,00065,  or  S  =  0",108 

Strength  of      The  comparative  strength  of  rectangular  beams 

fere^n^ngths  &  °^  cantilevers  of  different  cross-sections  and  spans, 

cross-sections.     but  of  same  materials  and  similarly  loaded  and  snp- 

ported,  is,  of  course,  directly  as  the  product  of  their  breadth  into  the 


62  SAFE    BUILDING. 

squares  of  their  depths,  divided  by  their  lengtli  of  span,  or 

x  =  ^  (32) 

Where  x  =  a,  figure  for  comparing  the  strength  of  different  beams 

of  same  material,  but  of  different  cross-sections  and  spans. 

Wliere  i=rtlie  breadth,  in  inclies. 

AVhere  d  =  thc  depth,  in  inches. 

Where  i  =  the  length  of  span,  in  feet. 

Stiffness  of      "^^^^  comparative  stiffness  or  amount  of  deflection 

beams  of   dif- of   different   rectangular   beams   or   cantilevers   of 
ferent  lensths&  ,     .      ,     ,     ,      n    ,        i  i 

cross-sections,     same  material,  and  similarly  loaded  and  supjiorted, 

but  of  different  cross-sections  and  spans,  would  be  directly  as  the 

cubes  of  their  respective  lengths,  divided  by  the  product  of  their 

respective  breadths  into  the  cubes  of  their  depths  or 

Where  a;  =  a  figure  for  comparing  the  amount  of  deflections  of 
beams  of  same  material  and  load,  but  of  different  spans  and  cross- 
sections. 

Where  L  =  the  length  of  span,  in  feet. 

Where  b  =  the  breadth,  in  inches. 

Where  (I  =  thc  depth  in  inches. 
Strength     and       ^^  ^^  '^  desired  to  calculate  a  wooden  girder  sup- 

deflection  of  iiorted  at  both  ends  and  to  carr\-  its  full  safe  uniform 

wooden    beams,  ,       ,         ,  ,   ^  , 

one  inch  thick,     load,  and  yet  not  to  deflect  enough  to  crack  plaster, 

the  following  will  simphfy  the  calculation  : 
TABLE  VIII. 


Spruce. 

Georgia  piue. 

"White  pine. 

White  oak. 

Hemlock. 

Calculattt  (x)  for 

transverse  strength 

only  if  d  is  greater  than 

HL 

L 

l-^L 

IIL 

\-t^ 

Calculate  (j-,)  for 
deflection  .mly 
if  d  is  less  than 

H^ 

L 

l^\L 

HL 

HL 

Where  L  =  the  length  of  span,  in  feet. 

Where  d  =  the  depth  of  beam,  in  inches. 

Where  a:  or  x,  =  is  found  according  to  Table  IX. 

To  find  the  safe  load  (.r)  or  (r,)  per  running  foot  of  span,  which  a 
beam  supported  at  both  ends,  and  1"  ihick  will  carry,  use  the  follow- 
ing table.     (Beams  two  inches  thick  will  safely  carry  twice  as  much 


WOODEN   BEAMS. 


63 


per  runnin<f  foot,  as  found  per  tabic,  beams  three  inches  thick  three 
times  as  mucb,  four  inches  tliick  four  times  as  much,  etc.) 


TABLE  IX. 


Spruce. 

Georgia 
pine. 

White  pine. 

White  oak. 

IHenilock. 

If  calculating 

for  transverse 

Streuglb    only 

use 

x=.ii(jy 

x=  133(1)' 

x=10o(z) 

x  =  122(0 

x=s-6{'^y 

If    calculating 

for  a  deflection 

(not    to    crack 

plaster)  use 

'■=Hir 

x.=r  133(9' 

..  =  95(j; 

^.  =  ioo(^) 

..=80(9" 

Where  a;  =  the  safe  load  in  lbs.,  per  running  foot  of  span,  wliich 
a  beam  one-inch  thick  will  carry  regardless  of  deflection,  if  supported 
at  both  ends,  and 

x,=:the  same,  but  without  deflecting  the  beam  enough  to 
crack  plaster;  for  thicker  beams  multiply  x  or  x,  by  breadth,  in 
inches. 

Calculate  for  either  x  or  x,  as  indicated  in  Table  VIII. 

Where  d=  the  depth  of  beam,  in  inches. 

Where  L  =  the  length  of  span  in  feet. 

If  a  beam  is  differently  supported,  or  not  uniformly  loaded,  also 
for  cantilevers,  add  or  deduct  from  above  result,  as  directed  in  cases 
1  to  8,  page  57. 

Example. 

Aflnor  of  19'  clear  span  is  to  he  built  with  spruce  beams,  to  carry 
100  lbs.  per  square  foot ;  what  size  beams  would  be  the  most  economical  f 

According  to  Table  VIII,  if  d=ll.  L  =  11.  19  =  221" 
•we  can  calculate  for  either  deflection  or  rupture  and  the  result  would 
be  the  same.  If  we  make  the  beam  deeper  it  will  be  so  stiff  that  it 
■will  break  before  deflecting  enough  to  crack  plastering  underneath  ; 
■while  if  we  make  the  beam  more  shallow  it  will  deflect  enough  to 
crack  plaster  before  it  carries  its  total  safe  load.  The  former  would 
be  more  economical  of  material,  but,  of  course,  in  practice  we  should 
certainly  not  make  a  wooden  beam  as  deep  as  22".  Whatever  depth 
we  select,  therefore,  less  than  22",  we  need  calculate  for  deflection 
only.    We  have,  then,  according  to  Table  IX,  second  column, 


54  SAFE   BUILDING. 

If  we  use  a  beam  12"  deep,  we  should  have 

193 

.- 95.^^3  =.24 

or  a  beam  l"x  12"  would  carry  24  lbs.  per  foot;    as  the  load  is  100 

lbs.  per  foot  we  should  need  a  beam  ^  =  4i  wide,  or  say  a  beam 

4"  X  12",  and  of  course  12"  from  centres. 
If  we  use  a  beam  14"  deep  we  should  have 

or  a  beam  1"  x  14"  would  carry  38  lbs.  per  foot,  we  need,  therefore, 
a  beam  of  width 

,  _100_^12" 
38  ~~"19 
or  we  must  use  a  beam  say  3"  x  14"  and  12"  from  centre,  or  a  beam 
4"xl4"  and  16"  from  centre.     For  if  the  beams  are  16"  from  cen- 
tres each  beam  will  carry  per  running  foot  1^.100  lbs.  =  133  lbs. 
and  a  4"  x  14"  will  carry  per  foot 
4.x,  =  4.38  =152. 

We  could  even  spread  the  beams  farther  apart,  except  for  the  dif- 
ficulty of  keeping  the  cross-furring  strips  sufficiently  stiff  for  lathing. 

Of  course  the  14"  beam  is  the  most  economical,  for  in  the  12" 
beam  we  use  4"  x  12"  =  48  square  inches  (cross-section)  of  material, 
and  our  beam  is  a  trifle  weak.  AVhile  with  the  14"  beam  we  use 
onlv  3"  X  14"  =  42  square  inches  of  material,  and  our  beam  has 
strength  to  spare.  The  4"  x  14"  beam  16"  from  centres  would  be 
just  as  strong  and  use  just  as  much  material  as  the  3"  x  14"  beam  12" 
from  centres.  If  we  wished  to  be  still  more  economical  of  material, 
we  might  use  a  still  deeper  beam,  but  in  that  case  it  would  be  less 
than  3"  thick  and  might  twist  and  warp.  If  the  beam  is  not  cross- 
bridged  or  supported  sideways  it  might  be  necessary  to  calculate  its 
strength  for  lateral  flexure.  That  it  will  not  shear  off  transversely 
we  can  see  readily,  as  the  load  is  so  light,  nor  is  there  much  danger 
of  longitudinal  shearing,  still  for  absolute  safety  it  would  be  better 
to  calculate  each  strain. 

Strength  of  col-  The  comparative  strength  of  columns  of  same  cross- 
len"ths!""^'^^"*  section  is  approximately  inversely  as  the  square  of 
their  len"-ths.     Thus,  if  x  be  the  strength  of  a  column,  whose  length 


PLATE    GIRDKUS. 


65 


is  Z.and  a;,  be  the  strength  of  a  column  whose  length  is  I„  then  we 
have  approximately 

x:x.=  L,2:i2,orx.=^'  (34) 

AVhere  a:,  =  approximately  the  strength  of  a  column,  L,  feet  long. 

AVherea;  =  the  strength  (previously  ascertained  or  known),  of  a 
column  of  same  cross-section,  and  L  feet  long. 

Where  L  and  i,=  the  respective  lengths  of  columns  in  feet. 

The  nearer  Z  and  A  are  to  each  other  the  closer  will  be  the  result. 

Strength  of  ool-      The  comparative  strength  per  square  inch  of  cross- 

umns    different        J-       of  columns  of  same  length,  but  of  different 

cross-sections.  ,     .     ,  .,       t         ^ 

cross-sections,  is,  approximately,  as  their  least  outside  diameter,  or 

side,  or 

,:,-l:l^,orx,  =  '^  (35) 

Where  x,  =  approximately  the  strength  of  a  column,  per  square 
inch,  whose  least  side  or  diameter  (outside)  is  =  b,. 

Where  a;  =  the  strength  per  square  inch  (previously  ascertained 
or  known)  of  a  column  of  same  length,  but  whose  least  side  or  diam- 
eter (outside)  is  =  b. 

The  more  f-'imilar  and  the  nearer  in  size  the  respective  cross-sec- 
tions are,  the  closer  will  be  the  result.  That  is,  the  comparison 
between  two  circular  columns,  each  1"  thick,  will  be  very  much 
nearer  correct  than  between  two  circular  columns,  one  f"  thick  and 
the  other  2"  thick,  or  between  a  square  and  a  circular  column.  The 
thicker  the  shell  of  a  column  the  less  it  will  carry  per  square  inch. 
The  formula  (34  and  35)  are  hardly  exact  enough  for  safe  practice, 
but  will  do  for  ascertaining  approximately  the  necessary  size  of  col- 
umn, before  making  the  detailed  calculation  required  by  formula  (3). 

The  approximate  thickness  required  for  the  Hanges  of  plate  gir- 
ders is  as  follows : 

Approximate  ^ 

thickness  of  ,/         "i                                           (2Q  i 

flange   of    plate  xz=- — =; ^     "^ 

girders.  0 

Where  a;=  the  approximate  thickness,  in  inches,  of  either  flange 
of  a  riveted  girder. 

Where  6  =  the  breadth  of  flange,  less  rivet  holes,  all  in  inches. 

AVhcre  r/=the  total  dejjth  of  girder  in  inches. 

Where  r  =  the  moment  of  resistance  in  inches. 

Wherea,  =  the  area  (less  rivet  holes)  of  cross-section  of  both 
angles  at  flange,  in  square  inches. 


fiA 


SAFE    BUILDING. 


w 


Ll 


4- 


-1- 


— ^ 


Fig.   1  I . 

materials,  and  are  as  follows 


DEFLECTION. 

The  derivation  of  the  following 
be  too  lengthy  to 
will  suffice  for  all 
practical  purposes  to  give  them. 
They  are  all  based  on  the  mod- 
uli  of    elasticitv  of   the    different 


--j-l-Xformulaj  would 
3*         go  into  here,  it 


FOR    A    CANTILEVER,    UNIFORMLY    LOADED. 
1      u3 


8  = 


e.x 


(37) 


FOR  A  CANTILEVER,  LOADED  AT  FREE  END. 


5 


O  1         ti\l^ 


-  C^-^:: 


Fig.  12. 

FOR   A   BEAM,   UNIFORMLY   LOADED. 


O 5       M.Z* 

38i*7t 


Fig.  13. 

FOR  A  BEAM,  LOADED  AT  CENTRE. 


(38) 


(39) 


If: 


I 


■Ik----' 


^\ 


<— 


-1-      H 

FIs.  14. 


48  "iJ 


(40) 


DEFLECTION. 


67 


FOR   A    BEAM,    LOADED    AT   ANY   POINT. 

Greatest  deflection  is  near  the  centre,  not  at  the  point  where  load 
is  applied.^ 

o w.m.n.(l-\-7i)  \fm.(l-\-n)  /^i\ 

^~        9.Le.i.         y         3 

Where  «  =  uniform  load,  in  pounds. 

Where  «J  =  concentrated  load,  in  pounds. 

Where  Z  =  length  of  span,  in  inches. 

Where  e=the  modulus  of  elasticity,  in  pounds-inch,  of  the  mate- 
rial, see  Tables  IV  and  V. 

Where  i  =  the  moment  of  inertia,  of  cross-section,  in  inches. 

Where  m  and  n  =  the  respective  distances  to  supports,  in  inches. 

Where  8=  the  greatest  xleflection,  in  inches  (see  Formulse  28  and 
29). 

FOR   A   CANTILEVER,    LOADED   AT   ANY  POINT. 

Greatest  deflection  is  at  free  end ;  if  y  =  distance  from  support  to 
load,  in  inches,  then  :  ^ 

3         e.i 

EXPANSION  AND  CONTRACTION  OF  MATERIALS. 
Expansion  of  All  long  iron  trusses,  say  about  eiglity  feet  long, 
Iron  trusses.  ^^  ^^^^.^  should  not  be  built-in  solidly  at  both  ends  ; 
otherwise  the  expansion  and  contraction  due  to  variations  of  the 
temperature  will  either  burst  one  of  the  supports,  or  else  cause  the 
truss  to  deflect  so  much,  as  to  crack,  and  possibly  endanger  the 
work  overhead.  One  end  should  be  left  free  to  move  (lengthwise 
of  truss)  on  rollers,  but  otherwise  braced  and  anchored,  tte  an- 
chor sliding  through  slits  in  truss,  as  necessary.  The  expansion 
of  iron  for  each  additional  single  degree  of  temperature,  Fahren- 
heit, is  about  equal  to  345Q0O  °^  '^^^  length,  that  is,  a   truss  145 

» The  point  of  greatest  deflection  can  never  be  further  from  the  centre  of  beam 
than  2-^0  of  the  entire  lei.glh  of  span.  It  can  as  a  rule,  therefore,  be  safely  as- 
sumed to  be  at  the  centre.    If  it  is  desired  to  find  its  exact  location,  use 

X=f^  («) 

where  /« —the  distance  from  weight  to  7!parer  support;  a;  =  the  distance  of  point 
of  greatest  liefleclion  from  farther  support;  and  ^  =  the  length  of  span;  x,  I  and 
wshouM  all  be  expressed  either  in  feet  or  inrhes. 
»  Formula  (42)  is  approximate  only,  but  sufficiently  exact  for  practical  use. 


68 


6AFK   BUILDI.N-G. 


feet  long  at  10°   Fahrenheit,  would  gain   in  length  (if  the  tempera- 

90.145  9 

ture  advanced  to  100°  Fahrenheit),  —  14-000  ^^  100  "^   ^  ^°"^'   °^' 

say,  lJ^inches,so  that  at  100°  Fahrenheit  the  truss  would  be  145  feet 
and  l^\y  inches  long;  this  amount  of  expansion  would  necessitate  roll- 
ers under  one  end.  Of  course  the  contraction  would  be  in  the  same 
proportion.  The  approximate  expansion  of  other  materials  for  each 
additional  degree  Fahrenheit  would  be  (in  parts  of  their  lengths),  as 
follows : 


Expansion  and  .  1 

contraction     of  ^  rought-iron  j^^qoo 

materials.  „    ^ .  1 

Cast-iron 


Steel 

Antimony 

Gold,  annealed. 

Bismutb 

Copper  

Brass 

Silver 

Gun  metal 

Tin 

Lead 

Solder 


162U0U 

1 
1510UO 

1 


166000 

1 
123000 

1 
130000 

1 
104000 

1 
95000 

1 
95000 

1 
90000 

1 

87000 

1 

•  63000 

1 

•  70000 


Pewter 

Platina 

Zinc 

Glass 

Granite 

Fire  Bricli 

Hard  Biick 

White  Marble. 

Slate 

Sandstone 

White  pine 

Cement 


1 

•  78000 

1 
208 jOO 

1 
62000 

1 
210000 

1 
208000 

1 
365000 

1 
600000 

1 
173000 

1 
173000 

1 
103000 

1 
440000 

1 
120000 


The  tension  due  to  each  additional  degree  of  Fahrenheit  would  be 
equal  to  the  modulus  of  elasticity  of  any  material  multiplied  by  the 
above  fraction  ;  or  about  186  pounds  per  square  inch  of  cross-section, 
for  wrought-iron.  Above  figures  are  for  linear  dimensions,  the 
superficial  extension  would  be  equal  to  twice  the  linear,  while  the 
cubical  extension  would  be  equal  to  three  times  the  linear. 

Water  is  at  its  maximum  density  at  about  39°  Fahrenheit ;  above 
tnat  it  expands  by  additional  heat,  and  below  that  point  it  expands 
by  less  heat.  At  32°  Fahrenheit  water  freezes,  and  in  so  doing  ex- 
pands nearly  r^  part  of  its  bulk,  this  strain  equal  to  about  30,000  lbs. 

per  square  inch  will  burst  iron  or  other  pipes  not  sufficiently  strong  to 
resist  such  a  pressure.  The  above  table  of  expansions  might  be  useful 
in  many  calculations  of  expansions  in  buildings;  for  instance,  were 


GRAPHICAL   ANALYSIS. 


69 


Fig.  15. 


we  to  make  the  sandstone  copings  of  a  building  in  10-foot  lengths, 

and  assume  the  variation  of  temperature  from  summer  sun  to  winter 

cold  would  be   about    150°  Fahrenheit,  each  etone  would  expand 

150.10         1     ,     ^  

103000  —  68      ^  ^°^'  °^'  ^^^'  about  J  inches,  quite  sutncient  to  open 

the  mortar  joint  and  let  the  water  in.  The  stones  should,  therefore, 
be  much  shorter. 

GRAPHICAL    METHOD    OF    CALCULATING     STRAINS.  —  NOTATION. 

Notation.      The  calculation  of  strains  in  trusses  and  arches  is 
1  based  on  the  law  Known  as  the 

"Parallelogram  of  Forces."  Be- 
fore going  in- 
to same  it  will 
be  necessary 
to  explain  the 
1  notation  used. 
If  Fig.  15  rep- 
resents a  truss, 
and  the  arrows 
the  loads,  and 

the  two  reactions  (or  supporting  forces),  we  should  call  the  left  reac- 
tion O  A  and  the  right  reaction  F  O.  The  loads  would  be,  taking 
them  in  their  order,  A  B,  B  C,  C  D,  D  E  and  E  F.  The  foot,  or 
lower  half,  of  left  rafter  would  be  called  B  K,  the  upper  half  C  I, 
while  the  respective  parts  of  right  rafter  would  be  G  E  and  II  D. 
The  King-post  (tie)  is  I  II,  and  the  struts  K  I  and  H  G,  while  the 
lower  ties  are  K  O  and  O  G. 

In  the  strain  diagram,  Fig.  16  (which  will  be  explained  presently), 
the  notation  is  as  usual ;  that  is,  loads  A  B,  B  C,  C  D,  etc.,  are  rep- 
resented in  the  strain  diagram  by  the  lines 
ab,  be,  cd,  etc.  Rafter  pieces  B  K,  C  I,  D  II 
and  E  G  are  in  the  strain  diagram  b  k,  c  i, 
dh  and  e  g  {g  and  k  falling  on  the  same  point). 
I  II  in  Fig  15  becomes  i  h  in  strain  diagram. 
K  I  becomes  k  i,  II  G  becomes  h  g,  O  K 
becomes  o  ^,  G  O  becomes  g  o,  O  A  becomes 
0  a  and  F  O  becomes /o. 

Or,  in  the  drawing  of  the  truss  itself  the 
lines  are  called,  not  by  letters  placed  at  the 
ends  of  the  lines,  but  by  letters  placed  each 
side  of  the  lines,  the  lines  being  between ;  it  is  also  usiial  to  put  these 


Fig.  I«. 


70 


SAFE    BUILDING. 


letters  in  capitals  to  distinguish  them  from  the  letters  representing 
the  strain  diagram,  which  are,  as  usual,  at  each  end  of  the  line  tliey 
represent. 

One  thing  is  very  important,  however,  and  that  is,  always  to  read 
the  pieces  off  in  the  correct  direction  and  in  their  proper  order.  For 
instance,  if  we  were  examining  the  joint  at  middle  of  left  rafter  we 
must  read  off  the  pieces  in  their  proper  order,  as  B  C,  C  I,  I  K, 
K  B,  and  not  jump,  as  B  C,  I  K,  C  I,  etc.,  as  this  would  lead  to  error. 
Still  more  important  is  it  to  read  around  the  joint  in  one  direction, 
as  from  left  to  right  (Fig.  17),  that  is,  in  the 
direction  of  the  arrow.  If  we  were  to  re- 
verse the  reading  of  the  pieces,  we  should 
find  the  direction  of  the  strain  or  stress  re- 
versed in  the  strain  diagram.  For  instance, 
if  we  read  K  I  and  then  find  its  correspond- 
ing line  k  i  in  the  strain  diagram,  we  find 
its  direction  downward,  that  is,  pulling 
awav  from  the  joint,  which  would  make  K  I  a  tie-rod,  which,  of 
course,  is  wrong,  as  we  know  it  is  a  strut.  If,  however,  we  had  read 
correctly  i  k\t  would  be  pushing  upwards,  which,  of  course,  is  correct 
and  is  the  action  of  a  strut. 

When  we  come  to  examine  the  joint  at  O,  however,  we  reverse  the 
above  and  here  have  to  read  k  i,  which  is  in  the  same  relative  direc- 
tion for  the  point  O,  as  was  i  k  for  the  point  at  centre  of  left  rafter. 


b' 

Fig.  18. 

The  arrows  in  the  accompanying  figure  (18)  show  how  each  joint 
must  be  read,  and  remember  always  to  read  the  pieces  in  their  proper 
succession. 

It  makes  no  difference  with  which  joint  or  with  which  piece  of  the 
joint  we  begin,  so  long  as  we  read  in  correct  succession  and  direc- 
tion, thus :  for  joint  No  1  we  can  read 
A  B,  B  K,  K  O  and  O  A 
or  K  O,  O  A,  A  B  and  B  K, 


PAIJALLELOGKAM    OF   FORCES. 


71 


or  B  K,  K  O,  0  A  and  A  B,  etc. 

In  the  strain  sheet  of  course  we  read  in  tlie  same  succession,  and 
it  will  be  found  that  the  lines,  as  read,  point  always  in  the  correct 
direction  of  the  strain  or  stress. 

PAKALLELOGKAM  OP  F0KCE9. 
Parallelogram  ^^  »  l^all  lying  at  the  point  A,  Fig.  19,  is  propelled 
of  Forces,  by  a  power  sufficient  to  drive  it  in  the  direction  of  B, 
and  as  far  as  B  in  one  minute,  and  at  B  is  again  propelled  by  a  power 
suflicient  to  drive  it  in  the  direction  of  and  as  far  as  the  point  C  in 
another  minute,  it  will,  of  course,  arrive  at  C  at  the  end  of  two  min- 
utes, and  by  the  route  ABC. 
If,  on  the  other  hand,  both  powers  had  been  applied  to  the  ball 
simultaneously,  while  lying  at  A,  Fig.  20, 
it  stands  to  reason  that  the  ball  would 
have  reached  C,  but  in  one  minute  and  by 
the  route  AC.  AC  (or  E  D),  is,  there- 
fore, called  the  resultant  of  the  forces  A  E 
V  and  D  A.  If,  now,  we  were  to  apply  to  the 
■"  ball,  while  at  A,  simultaneously  with  the 
forces  D  A  and  A  E,  a  third  force  (E  D) 
sufficient  to  force  the  ball  in  the  oppo- 
site direction  to  A  C  (that  is,  in  the  direc- 
tion of  C  A),  a  distance  equal  to  C  A  in 
one  minute  it  stands  to  reason  that  the 
ball  would  remain  perfectly  motionless  at  A,  as  C  A  being  the  result- 
ant (that  is,  the  result)  of  the  other  two  forces,  if  we  oppose 
them  with  a  power  just  equal  to  their  own 
result,  it  stands  to  reason  that  they  are  com- 
pletely neutralized.  Now,  applying  this  to 
a  more  practical  case,  if  we  had  tAvo  sticks 
lying  on  A  E  and  D  A,  Fig.  21,  and  holding 
the  ball  in  place,  and  we  apjjly  to  the  ball  a 
force  E  D  =  C  A  and  in  the  direction  C  A, 
we  can  easily  find  how  much  each  stick  must 
resist  or  push  against  the  ball.  Draw  a  line 
e  d,  Fig.  22,  parallel  to  E  U,  and  of  a  length 
at  any  convenient  scale  equal  in  amount  to/-^ 
force  E  D ;  through  e,  Fig.  22,  draw  a  e  par- 
allel to  A  E,  and  through  d  draw  (/  a  parallel  to  D  A,  then  the  tri- 
angle eda  (not  ead)  is  the  strain  diagram  for  ihe  Fig.  21,  and  d  a, 


Fig.   19. 


> 


Fig.  20. 


72 


SAFE   BUILDING. 


measured 


/ 


■e' 


by  the   same  scale   as  e  d,   is  the   amount  of  force   re- 

0  quired  for  the  stick  D  A  to  exert,  while  a  e, 

1  ^^  measured  by  the  same  scale,  is  the  amount  of 
/       ^\        force  required  for  the  stick  A  E  to  exert.    If 

'  \  in  place  of  the  force  E  D  we 

had   had    a    load,  the    same 
truths  would  hold  good,  but 
/'■'     /        we  should  represent  the  load 
•  '         /  by  a  force  acting  downward 

Q<''  /  in  a  vertical  and  plumb  line. 

\         /  Thus,  if  two  sticks,  B  A  and 

\  '  A  C,  Fig.  23,  are  supporting 

A  a  load  of  ten  pounds  at  their 

Figs.  21  and  22.  summit,  and   the   inclination 

of  each  stick  from  a  horizontal  line  is  45°,  we  proceed  in  the  same 
manner.  Draw  c  6,  Fig.  24,  at  any  scale  equal  to  ten  units,  through 
h  and  c  draw  h  a  and  a  c  at  angles  of  45°  each,  with  c  b,  then  meas- 
ure the  number  of  (scale  measure)  units  in  6  a  and  a  c,  which,  of 
course,  we  find  to  be  a  little  over  seven.  Therefore,  each  stick  must 
resist  with  a  force  equal  to  a  little  over  seven  pounds. 

Now,  to  find  the  di- 
rection of  the  forces. 
In  Fig.  23  we  read 
CB,BAand  AC,the 
corresponding  parts 
in  the  strain  diagram, 
Fig.  24,  are  c  b,  h  a 
and  a  c.  Now  the 
direction  of  c  6  is 
downwards,  therefore 
C  B  acts  downwards, 
which  is,  of  course, 
the  effect  of  a  weight.  •"'£«■  23  and  24. 

The  direction,  however,  of  6  a  and  a  c  is  upwards,  therefore  B  A  and 
A  C  must  be  pushing  upwards,  or  towards  the  weight,  and  therefore 
they  are  in  compression.  The  same  truths  hold  good  no  matter  how 
many  forces  we  have  acting  at  any  point;  that  is,  if  the  point  remains 
in  equilibrium  (all  the  forces  neutralizing  each  other),  we  can  con- 
struct a  strain  diagram  which  will  always  be  a  chimed  polygon  with 
as  many  sides  as  there  are  forces,  and  each  side  equal  and  parallel 
to  one  of   the  forces,  and  the  sides  being   in  the  same  succession 


ANALYSIS    OF   TRUSS. 


73 


Roof  Trusses,  simple  truss. 


Fig.  25. 


to  each  other  as  the  forces  are.     We  can  now  proceed  to  dissect  a 

Take  a  roof  truss  with  two  rafters 
and  a  single  tie-beam. 
The  rafters  are  sup- 
posed to  be  loaded  uni- 
formly, and  to  be  strong 
enough  not  to  give  way 
transversely,  but  to 
*v  transfer  safely  one-half 
\Cp^3,'  l^of  the  load  on  each  rafter 
to  be  supported  on  each 
joint  at  the  ends  of  the 
rafter.  Wo  consider 
each    joint    separately. 

Take  joint  No.  1,  Fig.  25. 

We  have  four  forces,  one  O  A  (the   left-hand  reaction),  being 

equal  to  half  the  load  on  the  whole  truss ;  next,  A  B,  equal  to  half 

the  load  on  the  rafter  B  E.     Then  we  have  the  force  acting  along 

B  E,  of  which  we  do  not  as  yet  know  amount  or  direction  (up  or 

down),  but  only  know  that  it  is  parallel  to  B  E;  the  same  is  all  we 

know,  as  yet,  of  the  force  E  O.  Now  draw,  at  any  scale,  Fig.  26, 

No.  1,  0  a  =:and  parallel  to 

O  A,  then  from  a  draw  a  b 

=  and  parallel  to  A  B  (a  6 

will,    of    course,    lap    over 

part  of  0  a,  but  this  does  not 

affect  anything).  Then  from 

b  draw  h  e  parallel  to  B  E, 

and  through  o  draw  e  o  par-  ^ 

allel  to  E  O.     Now,  in  read-|^ 

ing  off  strains,  begin  at  O  A, 

then  pass  in   succession   to 

A  B,  B  E  and  E  O.   Follow 

on  the  strain  diagram  Fig. 

26,   No.  1,  the  direction  as 

read  off,  with  the  finger  (that  j 

is,  0  a,  a  h,  b  e  and  e  o),  and  -i^ 

we  have   the   actual  direc- 
tions of  the  strains.     Thus  o  a  is  up,  therefore  pushing  up;   a  6  is 

down,  therefore  pushing  down;  &  e  is  downwards,  therefore  pushing 

against  joint  No.  1  (and  we  know  it  is  compression) ;  lastly,  e  o  is 


74 


BAFE   BUILDING. 


pushing  to  the  right,  therefore  puhing  away  from  the  joint  No.  1, 
and  -we  know  it  is'  a  tie-rod.  In  a  similar  manner  we  examine  tlie- 
joints  2  and  3,  getting  the  strain  diagrams  No.  2  and  No.  3  of  Fig. 
26.  In  Fig.  27,  we  get  tlie  same  results  exactly  as  in  the  above  three 
diagrams  of  Fig.  26,  only  for  simplicity  they  are  combined  into  one 
diagram.  If  the  single  (combination)  diagram.  Fig.  27,  should 
prove  confusing  to  the  student,  let  him  make  a  separate  diagram  for 
each  joint,  if  he  will,  as  in  Fig.  26.  Tlie  above  gives  the  principle 
jQ  of  calculating  the  strength  of  trusses,  graphically, 

and  will  be  more  fully  used  later  on  in  practical 

examples. 

Should  the  student  desire  a  fuller  knowledge  of 

the  subject,  let  him  refer  to  "  Greene's  Analysis  of 

Roof  Trusses,"  which  is  simple,  short,  and  by  far 

the  best  manual  on  the  subject. 

In  roof  and  other  trusses  the  line 
Line  of  Pressure    ^  ,       •  n     i 

Central.        of  pressure  or  tension  will  always 

be  co-incident  with  the  central  line  or  longitudinal 

axis  of  each  piece.     Each  joint  should,  therefore, 

be  so  designed  that  the  central  lines  or  axes  of  all 

the  pieces  will  go  through  one  iwint.     Thus,  for  in- 

Btance,^he  foot  of  a  king  post  should  be  designed  as  per  Fig.  28. 
In  roof-trusses  where  the  rafters  support  purlins,  the  rafters  must 

not  only  be  made  strong  enough  to  resist  the  compressive  strain  on 


Fig.  27. 


N^ 


^   M    i  y  ^ 7 


them,  but  in  addition  to  this 

enough  material  must  be  added 

to  stand  the  transverse  strain. 

Each   part  of    the    rafter   is 

treated  as  a  separate  beam, 

supported  at  each  joint,  and 

the  amount  of  reaction  at  each 

joint  must  be  taken  as  the  load 

at  the  joint.     The  same  holds 

good  of  the  tie-beam,  when  it  '^* 

has  a  ceiling  or  other  weights  suspended  from  it ;  of  course  these 

weights  must  all  be  shown  by  arrows  on  the  drawing  of  the  truss, 

so  a°s  to  get  their  full  allowance  in  the  strain  diagram.    Strains  in 

opposite   directions,  of  course,  counteract  each  otlier ;   the  stress, 

therefore,  to  be  exerted  by  the  material  need  only  be  equal  to  the 

difference  between  the  amounts  of  the  opposing  strains,  and,  of  course, 

this  stress  will  be  directed  against  the  larger  strain. 


ANALYSIS    OF    ARCH. 


76 


Fig.  29. 


We  consider  an 
arch  as  a  truss  with 
a  8 ucces sion  of 
straight  pieces; 
w  e  can   calculate 
it  graphically  the 
same  as  any  other 
■"""''1°'^ truss,  only  we  will 
find    that  the  ab- 
sence of  central  or 
inner     members 
(struts    and   ties) 
"will  force  the  line 
of  pressure,  as  a 
rule,  far  away  from 
the  central  axis.     Thus,  if  in  Fig.  29,  we  con- 
sider A  B  C  D  as    a    loaded    half-arch,  we 
know  that  it  is  held  in  place  by  three  forces, 
viz.: 

1.  The  load  B  C  L  M  which  acts  through  its  centre  of  gravity  as 
indicated  by  arrow  No.  1. 

2.  A  horizontal  force  No.  2  at  the  crown  C  D,  which  keeps  the 
arch  from  spreading  to  the  right. 

3.  A  force  at  the  base  B  A  (indicated  by  tlie  arrow  No.  3),  which 
keeps  the  arch  from  spreading  at  the  base.  Now  we  know  the  direc- 
tion and  amount  of  No.  1,  and  can  easily  find  Nos.  2  and  3.  In  an 
arch  lightly  loaded,  No.  2  is  always  assumed  to  act  at  two-thirds  way 
down  C  D,  that  is  at  F  (where  CE=:EF  =  FD  =  |C  D).  In 
an  arch  heavily  loaded,  No.  2  is  always  assumed  to  act  one-third  way 
down  C  D,  that  is  at  E;  further  the  force  No.  3  is  always  assumed 
to  act  through  a  point  two-thirds  way  down  B  A,  that  is  at  H  (where 
BG  =  GII=:HA=:^B  A).  The  reason  for  these  assumptions 
need  not  be  gone  into  here.  Therefore  to  find  forces  Nos,  2  and  3 
proceed  as  follows :  If  the  arch  is  heavily  loaded,  draw  No.  2  hori- 
zontally through  E  (C  E  being  equal  to  ^  C  D),  prolong  No  2  till  it 
intersects  No.  1  at  O,  then  draw  O  H  (H  A  being  equal  to  ^  B  A), 
•which  gives  the  direction  of  the  resistance  No.  3.  We  now  have  the 
three  forces  acting  on  the  arch  concentrated  at  the  point  O,  and  can 
easily  find  the  amounts  of  each  by  using  the  parallelogram  of  forces. 
Make  O  I  vertical  and  (at  any  scale)  equal  to  whole  load  (or  No.  1), 
draw  I  K  horizontally,  till  it  intersects  O  II  at  K;  then  scale  I  K, 


76 


SAFE   BUILDING. 


and  K  O  (at  same  scale  as  O  I),  which  will  give  the  amount  of  the 
forces  Nos.  2  and  3.  The  line  of  pressure  of  this  arch  A  B  C  D,  is 
Line  of  pressure  therefore  not  through  the  central  axis,  but  along 

not  central.  E  O  H  (a  curve  drawn  through  E  and  H  with  the 
lines  Nos.  2  and  3  as  tangents  is  the  real  line  of  pressure). 

Now  let  in  Fig.  30,  A  B  C  E  F  D  A  represent  a  half-arch.  We 
can  examine  A  B  C  D  same  as  before,  and  obtain  I  K  =  to  force 


Fig.  30. 
,    ,         ,  .        No.  2;  KO  =  to  resistance  (and  direction  of 

rfe^'^s..^  I  /  same)  at  U,  where  U  C  =  J  C  D;OI  being 
equal  to  the  load  on  D  A.  Now  if  we  consider 
the  whole  arch  from  A  to  F,  we  proceed  similar- 
ly. L  G  is  the  neutral  axis  of  the  whole  load 
from  A  to  F,  and  is  equal  to  the  whole  load,  at 
same  scale  as  O  I.     That  L  G  passes  through  D  is  accidental. 

Make  E  M  =  ^  E  F  and  draw  L  M ;  also  G  H  horizontally  till  it 
intersects  L  M  at  H,  then  is  G  II  the  horizontal  force  or  No.  2.  AVe 
now  have  two  different  quantities  for  force  No.  2,  viz. :  I  K  and  G  H, 
I  K  in  this  case  being  the  larger.  It  is  evident  that  if  the  whole  half- 
arch  is  one  homogeneous  mass,  that  the  greatest  horizontal  thrust  of 
any  one  part,  will  be  the  horizontal  thrust  of  the  whole,  we  select 
therefore  the  larger  force  or  I K  as  the  amount  of  the  horizontal  thrust. 
Now  make  S  P  =  to  I K  and  P  Q  =  to  No.  1 ,  or  load  on  A  D  and  P  R 
=  to  L  G  or  whole  load  on  F  A,  at  any  scale,  then  draw  Q  S  and  R  S. 
Now  at  O  we  have  the  three  forces  concentrated,  which  act  on  the 
part  of  arch  A  B  C  D,  viz. :  Load  No.  1  (=  P  Q),  horizontal  force 
No  2  (=  S  P)  and  resistance  K  O  (=  Q  S).  Now  let  No.  3  repre- 
sent the  vertical  neutral  axis  of  the  part  of  whole  load  on  F  D,  then 


LIXE    OF   TRESSURE.  77 

prolong  K  0  until  it  intersects  No.  3  at  T;  then  at  T  we  have  the 
three  forces  acting  on  the  part  of  arch  E  CDF,  viz. :  The  load  No. 
8  (=  Q  11),  the  thrust  from  A  B  C  D,  viz. :  O  T  (=  S  Q),  and  the 
resistance  N  T  (=  R  S).  To  obtain  N  T  draw  through  T  a  line  par- 
allel to  R  S,  of  course  R  S  giving  not  only  the  direction,  but  also  the 
amount  of  the  resistance  N  T.  The  line  of  pressure  of  this  arch 
therefore  passes  along  P  O,  O  T,  T  N.  A  curve  drawn  through 
points  P,  U  and  N  —  (that  is,  Avhere  the  former  lines  intersect  the 
joints  A  B,  D  C,  F  E)  —  and  with  lines  P  O,  O  T  and  T  N  as  tan- 
gents is  the  real  line  of  pressure.  Of  course  the  more  parts  we  divide 
the  arch  into,  the  more  points  and  tangents  will  we  have,  and  the 
nearer  will  our  line  of  pressure  approach  the  real  curve. 

Now  if  this  line  of  pressure  would  always  pass  through  the  exact 
centre  or  axis  of  the  arch,  the  compression  on  each  joint  would  of 
course  be  uniformly  spread  over  the  whole  joint,  and  the  amount  of 
this  compression  on  each  square-inch  of  the  joint  would  be  equal  to 
the  amount  of  (line  of)  pressure  at  said  joint,  divided  by  the  area  of 
the  cross-section  of  the  arch  in  square  inches,  at  the  joint,  but  this 
rarely  occurs,  and  as  the  position  of  the  line  of  pressure  varies  from 
the  central  axis  so  will  the  strains  on  the  cross  section  vary  also. 

Stress  at  intra-      Let  the  line  A  B  in  all  the  followinf  fio-ures  renre- 
dos  and  extra-        .    .,  ,.         r  .  ■         e  "  ,  ^,  ,        , 

dos.  sent  tlie  section  of  any  joint  of  an  arch  (the  thick- 

ness of  arch  being  overlooked)  C  D  the  amount  and  actual  position 
of  line  of  pressure  at  said  joint  and  the  small  arrows  the  sti-ess  or 
resistance  of  arch  at  the  joint. 

We  see  then  that  when  C  D  is  in  the  centre  of  A  B,  Fig.  31,  the 

stress  is  uniform,  that  is   the  joint  is,  uniformly  compressed,  the 

> — V  amount  of  compression 

(C  )  being  equal  to  the  aver- 

1^^  age  as  above.      As  the 

I  line  of  pressure  C  D  ap- 

A  Id      .  fe    Proachos  one  side,  Fig. 

illttttttttt  t  M  tft  tit  tttttti       ^^'  ^^^^  amount  of  com- 
Pjg^  3 1^  pression  on  that  side  in- 

creases, while  on  the  fur- 


© 


ther  side  it  decreases, 
until  the  line  of  pressure 
CD,  reaches   one-third 

Fig.  32.  see  there  is  no  compres- 


78 


SAFE    BUILDING. 


A  •  *♦■»♦! tttt ft t 


I 


h 


sion  at  A,  but  at  B  the  compression  is  equal  to  just  double  the  average 
as  it  was  in  Fig.  31.  Now,  as  C  D  passes  beyond  the  central  third  of 
'^ilN  A  B,  Fig.  34,  the  com- 

'  pression   at   the   nearer 

side  increases  still  fur- 
ther, while  the  further 
side  begins  to  be  sub- 
jected to  stress  in  the 
opposite  direction  or  ten- 
sion, this  action  increas- 
ing of  course  the  further 
C  D  is  moved  from  the 
t  r  t  t  L)  central  third.  This 
means  that  the  edge  of 
arch  section  at  B  would 
be^subject  to  very  severe 
crushing,  while  the  other 
When  the  line  C  D 


ng.  33. 


AiUi 


I 


n 


Fig.  34. 

edge  (at  A)  would  tend  to  separate  or  open, 
passes  on  to  the  edge  B,  the  nearer  two-thirds  of  arch  joint  will  be 
in  compression,  and  the  further  third  in  tension.  As  the  line  passes 
out  of  joint,  and  further  and  further  away  from  B,  less  and  less  of  the 
joint  is  in  compression,  while  more  and  more  is  in  tension,  until  the  line 
of  pressure  C  D  gets  so  far  away  from  the  joint  finally,  that  one-half 
of  the  joint  would  be  in  tension,  and  the  other  half  in  compression. 

Tension  means  that  the  joint  is  tending  to  open  upwards,  and  as 
arches  are  manifestly  more  fit  to  resist  crushing  of  the  joints  than 
opening,  it  becomes  apparent  why  it  is  dangerous  to  have  the  line  of 
pressure  far  from  the  central  axis.  Still,  too  severe  crushing  strains 
must  be  avoided  also,  and  hence  the  desirability  of  trying  to  get  the 
line  of  pressure  into  the  inner  third  of  arch  ring,  if  possible. 

But  the  fact  of  the  line  of  pressure  coming  outside  of  the  inner  third 
of  ai'ch  ring,  or  even  entirely  outside  of  the  arch,  does  not  necessarily 
mean  that  the  arch  is  unstable ;  in  these  cases,  however,  we  must  cal- 
culate the  exact  strains  on  the  extreme  fibres  of  the  joint  at  both  the 
inner  and  outer  edges  of  the  arch  (intrados  and  extrados),  and  see  to 
it  that  these  strains  do  not  exceed  the  safe  stress  for  the  material. 

The  formulae  to  be  used,  are  : 

For  the  fibres  at  the  edge  nearest  to  the  line  of  pressure 

And  for  the  fibres  at  the  edge  furthest  from  the  line  of  pressure 


PRESSUKE  ON  JOINT. 


79 


a  a.d  ^     ' 

Where  v  =  the  stress  in  lbs.,  required  to  be  exerted  by  the  extreme 
edge  fibres  (at  intrados  and  extrados). 

Where  x  =  the  distance  of  line  of  pressure  from  centre  of  joint  in 
inches. 

Where  a=:the  area  of  cross  section  of  arch  at  the  joint,  in  square 
inches. 

Where  p  =  the  total  amount  of  pressure  at  the  joint  in  lbs. 

Where  cf  =  the  depth  of  arch  ring  at  the  joint  in  inches,  measured 
from  intrados  to  extrados. 

When  the  result  of  the  formulae  (44)  and  (45)  is  a  positive  quan- 
tity the  stress  v  should  not  exceed  (y\  that  is  the  safe  compressive 
stress  of  the  material.  When,  however,  the  result  of  the  formula  (45) 
yields  a  negative  quantity,  the  stress  v  should  not  exceed  ('->  Y.that 
is  the  safe  tensile  stress  nf  the  material,  or  mortar. 

The  whole  subject  of  arches  will  be  treated  much  more  fully  later 
on  in  the  chapter  on  arches. 

A 


Fig.  35. 


80  SAFE    BUILDING. 

TO    ASCERTAIN    AMOUNT    OF    LOADS. 

Let  A  B  C  D  be  a  floor  plan  of  a  building,  A  B  and  C  D  are  the 
walls,  E  and  F  the  columns,  with  a  girder  between,  the  other  lines 
being  floor  beams,  all  12"  between  centres;  on  the  left  side  a  well- 
hole  is  framed  2'  x  2'.  Let  the  load  assumed  be  100  pounds  per 
square  foot  of  floor,  which  includes  the  weight  of  construction.  Each 
Load  on  ^^  *'^®  right-hand  beams,  also   the  three  left-hand 

Beams,  beams  E  L,  K  E  and  F  Q  will  each  carry,  of  course, 
ten  square  feet  of  floor,  or 

10.100=  1000  pounds  each  uniform  load.  Each  will  transfer  one- 
half  of  this  load  to  the  girder  and  the  other  half  to  the  wall.  Tlie 
tail  beam  S  N  will  carry  8  square  feet  of  floor,  or 

8.100  =  800  pounds  uniform  load.  One-half  of  this  load  will  be 
transferred  to  the  wall,  the  other  half  to  the  header  R  T,  which 
will  therefore  carry  a  load  of  400  pounds  at  its  centre,  one-half  of 
which  will  be  transferred  to  each  trimmer. 

The  trimmer  beam  G  M  carries  a  uniform  load,  one-half  foot  wide, 
its  entire  length,  or  fifty  pounds  a  foot  (on  the  off-side  from  well- 
hole),  or 

50.10=:  500  pounds  uniform  load,  one-half  of  which  is  transferred 
to  the  girder  and  the  other  half  to  the  wall.  The  trimmer  also 
carries  a  similar  load  of  fifty  pounds  a  foot  ou  the  well-hole  side,  but 
only  between  M  and  R,  which  is  eight  feet  long,  or 

50.8  =  400  pounds,  the  centre  of  this  load  is  located,  of  course, 
half  way  between  M  and  R,  or  four  feet  from  support  M,  and  six 
feet  from  support  G,  therefore  M  will  carry  (react) 

=  240  pounds  and  G  wiU  carry 


10 
4.400 


160  pounds. 


10 

See  FormuljE  (14)  and  (15). 

We  also  have  a  load  of  200  pounds  at  R,  transferred  from  the 
header  on  to  the  trimmer ;  as  R  is  two  feet  from  G,  and  eight  feet 
from  M,  we  will  find  by  the  same  formulae,  that  G  carries 
8.200 


10 
2.200 


=  IGO  pounds  and  M  carries 
=  40  pounds. 


10 
So  that  we  find  the  loads  which  the  trimmer  transfers  to  G  and  M, 

as  follows : 

At  M=250  +  240-j-  40  =  530  pounds. 

«  G  =  250 +  160 +  160  =  5 70  pounds. 


DISTIilBUTION   OF   LOADS. 


81 


The  loads  which  trimmer  O  I  transfers  to  wall  and  girder  will,  of 
Load  on  course,  be  similar.    We  therefore  find  the  total  load- 

Walls,     ing^  as  follows: 
On  the  wall  A  B  : 

At  L  =  500  pounds. 
"  M=  530  pounds. 
"  N=  400  pounds. 
«  O  =  530  pounds. 
"  P  =  500  pounds. 
"  Q  =  _500  pounds. 
Total  on  wall  A  B  ==  2'JCO  pounds. 

On  the  wall  C  D  we  have  six  equal  loads  of  600  pounds  each,  a 
Load  on  Girder,  total  of  3000  pounds. 


II 


O 

o 

I 


o 

0 

II 


o 


o 
o 
o 


I 

I 


o 
o 
o 

It 


i  '  '  I 

I'o— i I     — 4'd"-       • 


W 


H 


If- 


— ji^   —  2^o-- 


o  --- 


_^.o-— 


f- 


I'O 


F 


-:>, 


f 


Fig.  36. 

On  the  girder  E  F,  we  have: 
At  E  from  the  left  side  500  pounds,  from  the 

right  500  pounds. 
At  G  from  the  left  side  570  pounds,  from  the 

right  500  pounds. 
At  II  from  the  left  side  nothing,  from  the  right 

500  pounds. 
At  I  from  the  left  side  570  pounds,  from  the 

right  500  pounds. 
At  K  from  the  left  side  500  pounds,  from  the 

right  500  pounds. 


Total  1000  pounds. 
Total  1070  pounds. 
Total  600  pounds. 
Total  1070  pounds. 
Total  1000  pounds. 


82 


SAFE    DUILDING. 


At  F  from  the  left  side  500 
right  500  pounds. 


pounds,  from  the 

Total  1000  pounds. 
Total  on  girder  5640  pounds. 
As  the  girder  is  neither  uniformly  nor  symmetrically  loaded,  we 
must  calculate  by  Formulse  (16)  and  (17),  the  amount  of  each  reac- 
tion, which  will,  of  course,  give  the  load  coming  on  the  columns  E 
and  F.    (These  columns  will,  of  course  carry  additional  loads,  from 
the  girders  on  opposite  side,  further,  the  weight  of  the  column  should 
be  added,  also  whatever  load  comes  on  the  column  at  floor  above.) 
Girder  E  F  then  transfers  to  columns, 

At  E  =  1000+  (f  1070)  +  (|.  500)  4-  (|.1070)  +  (4.  1000)  -}- 
(0.  1000)  =  2784  pounds. 

At  F  =  1000  +  (|.  1000)  +  (f .  1070)  +  (|.  500)  +  (i.  1070)  + 
(0.  1000)  =  2856  pounds. 

As  a  ciieck  the  loads  at  E  and  F  must  equal  the  whole  load  on  the 
girder,  and  we  have,  in  effect, 

2  784-}- 2856  =  5640. 
Now  as  a  check  on  the  whole  calculation  the  load  on  the  two  col- 
umns and  two  walls  should  equal  the  whole   load.     The  whole  load 
being  20'x6'x  100  pounds  minus  the  well-hole  2'x2'x  100  pounds, 
or  12000  —  400  =  11600  pounds. 
And  we  have  in  effect. 

Load  on  A  B  =    2960  pounds. 

"CD  =    3000  pounds. 

"       two  columns  =    5640  pounds. 

Total  loads  =11600  pounds. 

We  therefore  can  calculate  the  strength  of  all  the  beams,  headers 

and  trimmers  and  girders,  with  loads  on,  as  above  eiven. 

For  the  columns  and  walls,  we  must 
however  add,  the  weight  of  walls  and 
columns  above,  including  all  the  loads 
coming  on  walls  and  columns  above  tlie 
point  we  are  calculating  for,  also  what- 
ever load  comes  on  the  columns  from 
the  other  sides.     If  there  are  openings 

...  ..    in  a  wall,  one-half  the 
Load  over  Wall 

openings.        load  over  each  opening 

goes  to  the  pier  each  side  of  the  opening, 

including,  of  course,  all  loads  on  the  wall 

above  the  opening. 

Thus  in  Fi"^ure  37,  the  weight  of  walls  would    be  distributed,  as 


Fi2.  37, 


WIND    AND    SNOW. 


83 


indicated  by  etched  linos;  where,  however,  the  opening  in  the  wall 
is  very  small  compared  to  the  mass  of  wall-space  over,  it  would,  of 
course,  be  absurd  to  consider  all  this  load  as  on  the  arch,  and  pract- 
ically, after  the  mortar  has  set,  it  would  not  be,  but  only  an  amount 
about  equal  to  the  part  enclosed  by  dotted  lines  in  Figure  38,  the 
inclined  lines  being  at  an  angle  of  60°  with  the  horizon.  Where 
only  part  of  the  wall  is  calculated  to  be  carried  on  the  opening,  the 
wooden  centre  should  be  left  in  until  the  mortar  of  the  entire  wall 
has  set.  In  case  of  beams  or  lintels  the  wall  should  be  built  up  until 
the  intended  amount  of  load  is  on  them,  leav- 
ing them  free  underneath;  after  the  intended 
load  is  on  them,  they  should  be  shored  up, 
until  the  rest  of  wall  is  built  and  thoroughly  set. 

Wind-pressure  on  a  roof  is 
Wind  Pressure  ,,  , 

and  Snow,      gcnerallyassumcdat  a  certain 

load  per  square  foot  superficial  measurement 
of  roof,  and  added  to  the  actual  (dead)  weight 
of  roof ;  except  in  large  roofs,  or  where  one 
foot  of  truss  rests  on  rollers,  when  it  is  im- 
portant to  assume  the  wind  as  a  separate  force, 
acting  at  right  angles  to  incline  of  rafter.  ^'^'  ^^' 

The  load  of  snow  on  roofs  is  generally  omitted,  when  wind  is 
allowed  for,  as,  if  the  roof  is  very  steep  snow  will  not  remain  on 
it,  while  the  wind  pressure  will  be  very  severe  ;  while,  if  the  roof  is 
flat  there  will  be  no  wind  pressure,  the  allowance  for  which  will,  of 
course,  offset  the  load  of  snow. 

If  the  roof  should  not  be  steep  enough  for  snow  to  slide  off,  a 
heavy  wind  would  probably  blow  the  snow  off. 

In  case  of  "  continuous  girders,"  that  is,  beams  or  girders  sup- 
ported at  three  or  more  points  and  passing  over  the  intermediate 
supports  without  being  broken,  it  is  usual  to  allow  moi'e  load  on  the 
central  suj)ports,  than  the  formulae  (14)  to  (17)  would  give.  This 
subject  will  be  more  fully  dealt  with  in  the  chapter  on  beams  and 
girders. 

FATIGUE. 

If  a  load  or  strain  is  applied  to  a  material  and  then  removed,  the 
material  is  supposed  to  recover  its  first  condition  (provided  it  has 
not  been  strained  beyond  the  limit  of  elasticity).  This  practically, 
however,  is  not  the  case,  and  it  is  found  that  a  small  load  or  strain 
often  applied  and  removed  will  do  more  damage  (fatigue  the  mate- 
rial more)  than  a  larger  one  left  on  steadily.    Most  loads  in  buildings 


84  SAFE   BUILDING. 

are  stationary  or  "  dead  "  loads.  But  where  there  are  "  moving  " 
loads,  such  as  people  moving,  dancing,  marching, 
°*  "Loads.  etc.,  or  machinery  vibrating,  goods  being  carted  and 
dumped,  etc.,  it  is  usual  to  assume  larger  loads  than  will  ever  be  imposed ; 
sometimes  going  so  far  as  to  double  the  actual  intended  load,  or  what 
amounts  to  the  same  thing,  doubling  (or  increasing)  the  factor-of-safety, 
in  that  case  retaining,  of  course,  the  actual  intended  load  in  the  calcu- 
lations. This  is  a  matter  in  which  the  architect  must  exercise  his 
judgment  in  each  individual  case. 


you^•DATION8.  85 


CHAPTER  II. 


FOUNDATIONS. 


Nature  of  Soils.  The  nature  of  the  soils  usually  met  with  on  build- 
in"'  sites  are:  rock,  gravel,  sand,  clay,  loamy  earth,  "made"  ground 
and  marsh  (soft  wet  soil). 

If  the  soil  is  hard  and  practically  non-compressible,  it  is  a  good 
foundation  and  needs  no  treatment ;  otherwise  it  must  be  carefully 
prepared  to  resist  the  weight  to  be  superimposed. 

The  base-courses  of  all  foundation  walls  must  be 
Stepping 

Courses,  spread  (or  stepped  out)  sufficiently  to  so  distribute  the 

weio-ht  that  there  may  be  no  appreciable  settlement  (compression) 

in  the  soil. 

Two  important  laws  must  be  observed :  — 

1.  All  base-courses  must  be  so  proportioned  as  to  produce  exactly 
the  same  pressure  per  square  inch  on  the  soil  under  all  parts  of 
buUdin"'  where  the  soil  is  the  same.  Where  in  the  same  building  we 
meet  with  different  kinds  of  soils,  the  base-courses  must  be  so  pro- 
portioned as  to  produce  the  same  relative  pressure  per  square  inch 
on  the  different  soils,  as  will  produce  an  equal  settlement  (compres- 
sion) in  each. 

2.  "Whenever  possible,  the  base-course  should  be  so  spread  that  its 
neutral  axis  will  correspond  with  the  neutral  axis  of  the  superim- 
posed weight ;  otherwise  there  will  be  danger  of  the  foundation  walls 
setthng  unevenly  and  tipping  the  walls  above,  producing  unsightly  or 
even  dangerous  cracks. 

Example. 

In  a  church  the  gable  wall  is  1'  G"  thick,  and  is  loaded  (including 
weight  of  all  walls,  floors  and  roofs  coming  on  same)  at  the  rate  of  b2 
lbs,  per  square  inch.  The  small  piers  are  12"  x  12"  and  5'  high,  and 
carry  a  floor  space  equal  to  14'  x  10'.  What  should  be  the  size  of  base- 
courses,  it  being  assumed  that  the  soil  will  safely  stand  a  pressure  of  30 
lbs.  per  square  inch  f 

If  "we  were  to  consider  the  wall  only,  we  should  have  the  total 
pressure  on  the  soil  per  running  inch  of  wall,  18.52  =  93G  lbs. 


86  SAFE   BUILDIXG. 

Dividing  this  by  30  lbs.,  tlie  safe  pressure,  we  should  need  -^  =: 

31,2"  or  say  32"  width  of  foundation,  or  we  should  step  out  each  side 

of  foundation  wall  an  amount  — — -  =:  7"  each  side. 

2 

Xow  the  load  on  pier,  assuming  the  floor  at  100  lbs.  per  square 
foot,  would  be  14  X  10  X  100  =  14000  lbs.  To  this  must  be  added 
the  weight  of  the  pier  itself.  There  are  5  cubic  feet  of  brickwork 
(weighing  112  pounds  per  foot)  =  5.11 2^500  lbs.,  or,  including  base- 
course,  a  total  load  of  say  15000  lbs.  This  is  distributed  over  an  av- 
erage of  144  square  inches ;  therefore  pressure  per  square  inch  under 
pier. 

15000       ,  „ .  ,rto  „ 

=  104  or,  say,  100  lbs. 

144 

We  must  therefore  make  the  foundation  under  pier  very  much 
wider,  in  order  to  avoid  unequal  settlements.  The  safe  pressure  per 
square  inch  Ave  assumed  to  be  30  lbs.;  therefore  the  area  requiisd 

would  be  =  — orp  ^=  500  square  inches,  or  a  square  about  22"  x  22". 
We  therefore   shall  have  to  step  out  each  side  of  the  pier   an 

amount  "^ =  5  . 

2 

The  safe  compressions  for  different  soils  are  given  in  Table  V,  but 
in  most  cases  it  is  a  matter  for  experienced  judgment  or  else  experi- 
ment. 

Testing  ^^  ^^  usual  to  bore  holes  at  intervals,  considerablj' 

Soils,  deeper  than  the  walls  are  intended  to  go,  at  some  s])Ot 
where  no  pressure  is  to  take  place,  thus  enabling  the  architect  to 
judge  somewhat  of  the  nature  of  the  soil.  If  this  is  not  sufficient,  he 
takes  a  crowbar,  and,  running  it  down,  his  experienced  touch  should 
be  able  to  tell  whether  the  soil  is  solid  or  not.  If  this  is  not  suffi- 
cient, a  small  boring-machine  should  be  obtained,  and  samples  of  the 
soil,  at  different  points  of  the  lot,  bottled  for  every  one  or  two  feet 
in  depth.  These  can  be  taken  to  the  office  and  examined  at  leisure. 
The  boring  should  be  continued  if  possible,  until  hard  bottom  is 
struck. 

If  the  ground  is  soft,  new  made,  or  easily  compressible,  experi- 
ment as  follows:  Level  the  ground  off,  and  lay  down  four  blocks 
each,  say  3"  x  3" ;  on  these  lay  a  stout  platform.  Alongside  of  jilai- 
form  plant  a  stick,  with  top  level  of  platform  marked  on  same.  Xow 
pile  weight  onto  platform  gradually,  and  let  same  stand.  As  soon  as 
platform  begins  to  sink  appreciabl}'  below  the  mark  on  stick,  )0U 


WET    FOUXDATIOXS. 


87 


have  the  practical  ultimate  resistance  of  the  foundation ;  this  divided 
by  36  gives  the  ultimate  resistance  of  the  foundation  per  square  inch. 
One-tenth  of  this  only  should  be  considered  as  a  safe  load  for  a  per- 
manent building. 
Drainage  of      I^rainage  is  essential  to  make  a  building  healthy,  but 

-oi'-  can  hardly  be  gone  into  in  these  articles.  Sometimes 
it  is  also  necessary  to  keep  the  foundations  from  being  undermined. 

It  is  usual  to  lead  off  all  surface  or  spring  water  by  means  of  blind 
drains,  built  underground  with  stone,  gravel,  loose  tile,  agricultural- 
tile,  half-tile,  etc.  To  keep  dampness  out,  walls  are  cemented  and 
then  asphalted,  both  on  the  outsides.  If  the  wall  is  of  brick,  the 
Damp-proof-  cementing  can  be  omitted.     Damp-courses  of  slate  or 

'"£•  asphalt  are  built  into  walls  horizontally,  to  keep  damp- 
ness from  rising  by  capillary  attraction.  Cellar  bottoms  are  con- 
creted and  then  asphalted ;  where  there  is  pressure  of  water  from 


c 


& 


I 


l^'^-  39.  Fig.  40. 

underneath,  such  as  springs,  tide-water,  etc.,  the  asphalt  has  to  be 
sufficiently  weighted  down  to  resist  same,  either  with  brick  paving 
or  concrete. 

Where  there  are  water-courses  they  should  be  diverted  from  the 
foundations,  but  never  dammed  up.  They  can  often  be  led  into  iron 
or  other  wells  sunk  for  this  purpose,  and  from  there  pumped  into  the 
building  to  be  used  to  flush  water-closets,  or  for  manufacturing  or 
other  purposes.  Clay,  particularly  in  vertical  or  inclined  layers,  and 
sand  are  the  foundations  most  dangerously  a£fected  by  water  as  they 
are  apt  to  be  washed  out. 

Where  a  very  wide  base-course  is  required  by  the  nature  of  the  soil, 
it  is  usual  to  step  out  the  wall  above  gradually ;  the  angle  of  stepping 
should  never  be  more  acute  than  60°,  or,  as  shown  in  Figure  39.  Care 
must  also  be  taken  that  the  stepped-out  courses  are  sufliciently  wide 
to  project  well  in  under  each  other  and  wall,  to  prevent  same  break- 
ing through  foundation,  as  indicated  in  Figure  40. 

Where,  on  account  of  party  lines  or  other  buildings,  the  stepping 


88 


SAFE   BUILDING. 


out  of  a  foundation  wall  has  to  be  done  entirely  at  one  side,  the  step- 
ping should  be  even  steeper  than  60°,  if  possible ;  and  particular  at- 
tention must  be  paid  to  anchoring  the  walls  together  as  soon  and  as 
thoroughly  as  possible,  in  order  to  avoid  all  danger  of  the  foundation 
wall  tipping  outwardly. 

Where  a  front  or  other  wall  is 
composed  of  isolated  piers,  it  i> 
well  to  combine  all  their  founda- 
tions into  one,and  to  step  the  pier? 

down  for  this  purpose,  as  shown  ^ 

in  Figure  41.    Where  there  is  not  F\i,.  4i. 

sufficient  depth  for  this  purpose,  inverted  arches  must  be  resorted  to. 

The  manner  of  calculating  the  strength  of  inverted  arches  will 

be  given  under  the  article  on  arches.    Inverted  arches  are  not  recom- 

_    .         mended,  however  (except  where  the  foundation  wall  is 
Inverted  '.  ,    ,,      ^         . 

arches,  by  necessity  very  shallow),  as  it  requires  great  care  and 

good  mechanics  to  build  them  well. 

Two  things  must  particularly  be  looked  out  for  :  1,  That  the  end 

arch  has  sufficient 
pier  or  other  abut- 
ment; otherwise  it 
will  throw  the  pier 
out,  as  indicated  in 
Figui-e  42.  (This 
will  form  part  of 
calculation  of 
Fig.  42.  sti-ength  of  arch.) 

"Where  there  is  danger  of  this,  ironwork  should  be  resorted  to,  to  tie 
,  ^,         back  the  last  pier. 

8lZ6  of  Sk6W* 

back,  2.  The  skew-back  of  the  arch  should  be  sufficiently 

wide  to  take  its  proportionate 
share  of  load  from  the  pier  (that 
is,  amount  of  the  two  skew-backs 
should  be  proportioned  to  balance 
of  pier  or  centre  jiart  of  pier,  as 
the  width  of  opening  is  to  width 
of  pier)  ;  otherwise  the  pier  would 
be  apt  to  crack  and  settle  past 
arch,  as  shown  in  Figure  43. 
An  easy  way  of  getting  the 
graphically  is  given  below.       In   Figare  44, 


width  of 


Fig.  43. 

skew-back 


INVERTED    ARCHES. 


89 


draw  A  B  horizontally  at  springlng-Iine  of  inverted  arch;  bisect 
A  C  at  F,  and  C  B  at  E.  Draw  E  O  at  random  to  vertical  through 
F ;  then  draw  0  C,  and  parallel  to  0  C  draw  G  D ;  then  is  C  D  the 
required  skew-back.* 

A  good  way  to  do 
is  to  give  the  arches 
wide  skew-backs,  and 
then  to  introduce  a 
thick  granite  or  blue, 
stone  pier  stone  over 
them,  as  shown  in 
Figure  45.  This  will 
force  all  down  evenly 
and  avoid  cracks. 
The  stone  must  be 
Fig.  44.  thick  enough  not  to 

break  at  dotted  lines,  and  should  be  carefully  bedded. 

Example. 

A  foundation  pier  carrying  150000  lbs.  is  5'  wide  and  3'  hroad 
The  inverted  arches  are  each  24"  deep.  What  thickness  should  the 
granite  block  have  ? 

W  e  have 
here  vh'tual- 
ly  a  granite 
beam,  60" 
long  and  36" 
broad,  s  u  p - 
ported  at  two 
points  (the 
centre  lines 
of  s  k  e  w- 
backs)  30" 
apart.     The 


\ 


Ffg.  45. 


load  is  a  uniform  load  of  150000  lbs.      The  safe  modulus  of  rupture, 
according  to  Table  V,  for  average  granite  is  ( -—  j  =  180  lbs. 


iln  reality  C  D  should  be  somewhat  larger  than  the  amount  thus  obtained;  but 
this  can  be  overlooked,  except  in  cases  where  the  pier  approaches  in  widtli  the 
width  of  opening.  In  such  cases,  however,  stepping  can  generally  be  resorted  to 
in  place  of  inverted  arches.  Then,  too,  if  the  opening  were  very  wide  and  the 
line  of  pressure  came  very  much  outside  of  central  third  of  C  D,  it  might  be 
necessai-y  to  still  f  ui-thcr  increase  the  width  of  skew-back,  C  D. 


90 


SAFE    BUILDING 


The  bending  moment  on  this  beam,  according  to  Formula  (21)  is 

;  675  000 


u.l       150  000.36 
m  =  -s-  = 5 


The  moment  of  resistance,  r,  is,  from  Formula  (18) 

m  G  75  000 


(r) 


180 


-=3  750 


From  Table  I,  No.  3,  we  find 
r=  -Q-,  therefore 


bd^ 


=  3750  ;  now,  as  6  =  36,  transpose  and  we  have 


(Z2  — 


3750.6 


36 


625. 


Therefore  d  =  25"  or  say  24". 
have  to  be  5'  x  3'  x  2'. 


Fig.  46. 


The  size  of  granite  block  would 


As  this  would  be  a  very  un- 
wieldy block,  it  might  be  split 
in  two  lengthwise  of  pier ;  that 
is,  two  stones,  each  5'  x  18"  x 
2'  should  be  used,  and  clamped 
together.  Before  building  piers, 
he  arch  should  be  allowed  to 
get  thoroughly  set  and  hardened, 
to  avoid  any  after  shrinkage  of 
the  joints. 


A  parabolic  arch  is  best.  Next  in  order  is  a  pointed  arch,  then 
a  semi-circular,  next  elliptic,  and  poorest  of  all,  a  segmental  arch,  if  it 
is  very  flat.  But,  as  before  mentioned,  avoid  inverted  arches,  if  pos- 
sible, on  account  of  the  difficulty  of  their  proper  execution. 

„     .  A  rock  foundation  makes  an  excellent  one,  and  needs 

Rock  , 

foundations.  UtQe  treatment,  but  is  apt  to  be  troublesome  because  of 

water.  Remove  all  rotten  rock  and  step  off  all  slanting  surfaces,  to  make 


•  Co/sciJETE: 

Fig.  47. 

level  beds,  filling  all  crevices  with  concrete,  as  shown  in  Figure  47 : 
In  no  case  build  a  wall  on  a  slanting  foundation. 
Look  out  for  springs  and  water  in  rock  foundations.     "\^Tiere  soft 


ROCK   FOUXDATIOXS.  91 

soils  are  met  in  connection  with  rock,  try  and  dig  down  to  solid  rock, 
or,  if  this  is  impossible,  on  account  of  the  nature  of  the  case  or  ex- 
pense, dig  as  deep  as  jwssihle  and  put  in  as  wide  a  concrete  base- 
course  as  possible.  If  the  l)ad  spot  is  but  a  small  one,  arch  over 
from  rock  to  rock,  as  shown  in  Fl"-ure  48  : 


Fig.  48. 

_      .  ,       Good  hard  sand  or  even  quicksand  makes  an  excel- 

Sand,  gravel  .        ,     .         ,.  '■    , 

and  clay,  lent  foundation,  if  it  can  be  kept  from  shifting  and 

clear  of  water.  To  accomplish  this  purpose  it  is  frequently  "  sheath- 
piled  "  each  side  of  the  base  course. 

Gravel  and  sand  mixed  make  an  excellent,  if  not  the  best  founda- 
tion ;  it  is  practically  incompressible,  and  the  driest,  most  easily 
drained  and  healthiest  soil  to  build  on. 

Clay  is  a  good  foundation,  if  in  horizontal  layers  and  of  sufficient 
thickness  to  bear  the  superimposed  weight.  It  is,  however,  a  very 
treacherous  material,  and  apt  to  swell  and  break  up  with  water  and 
frost.  Clay  in  inclined  or  vertical  layers  cannot  be  trusted  for  im- 
portant buildings,  neither  can  loamy  earth,  made  ground  or  marsh. 
If  the  base-course  cannot  bo  sufficiently  spread  to  reduce  the  load  to 
a  minimum,  pile-driving  has  to  be  resorted  to.  This  is  done  in  many 
Short  different  ways.     If  there  is  a  layer  of  hard  soil  not  far 

piles,  down,  short  piles  are  driven  to  reach  down  to  same. 
These  should  be  of  sufficient  diameter  not  to  bend  under  their  load ; 
they  should  be  calculated  the  same  as  columns.  The  tops  should 
be  well  tied  together  and  braced,  to  keep  them  from  wobbling  or 
spreading. 

Example. 

Georrjia-pine  piles  of  IG"  diameter  are  driven  through  a  layer  of 
soft  soil  15'  deep,  until  they  rest  on  hard  bottom.  What  will  each  pile 
safely  carry  f 

The  pile  evidently  is  a  circular  column  15'  long,  of  IG"  diameter, 
solid,  and  we  should  say  with  rounded  ends,  as,  of  course,  its  bear- 


92 


SAFK   BUILDING. 


ings  cannot  be  perfect.    From  Formula  (3)  we  find,  then,  that  the 
pile  will  safely  carry  a  load. 


a. 
w  =  — 


/  c  \ 

-^j2j^f  now  from  Table  I,  Section  No.  7,  and  fifth 


1+  ^ 
column,  we  have 

22     „       22.82 

From  the  same  table  we  find,  for  Section  No.  7,  last  column, 

0  t^  82 

From  Table  IV  we  find  for  Georgia  pine,  along  fibres,  ^■4^  =  750. 
And  from  Table  11,  for  wood  with  rounded  ends, 
n  =  0,00067,  therefore : 

201.750 150750 

^~.    ,    1802.0,00067  ~  "2;357~  ~  ^^958, 
•■  16 

or  say  SO  tons  to  each  pile. 

gj,^^  Sometimes  large  holes  are  bored  to  the  hard  soil  and 

piles,  filled  with  sand,  making  "  sand  piles."  This,  of  course, 
can  only  be  done  where  the  intermediate  gi-ound  is  sufliciently  firm  to 
keep  the  sand  from  escaping  laterally'.  Sometimes  holes  are  dug  down 
and  filled  in  with  concrete,  or  brick  piers  are  built  down ;  or  large  iron 
cylinders  are  sunk  down  and  the  space  inside  of  them  driven  full  of 
piles,  or  else  excavated  and  filled  in  with  concrete  or  other  masonry, 
or  even  sand,  well  soaked  and  packed.  If  filled  with  sand,  there  should 
first  be  a  layer  of  concrete,  to  keep  the  sand  from  possibly  escaping 
at  the  bottom. 

Where  no  hard  soil  can  be  struck,  piles  are  driven  over  a  large 
area,  and  numerous  enough  to  consohdate  the  ground ;  they  should 
not  be  closer  than  two  feet  in  the  clear  each  way,  or  they  will  cut  up 
the  ground  too  much.  The  danger  here  is  that  they  may  press  the 
ground  out  laterally,  or  cause  it  to  rise  where  not  weighted.  Some- 
times, by  sheath-piling  each  side,  the  ground  can  be  sufficiently  com- 
pressed between  the  piles,  thereby  being  kept  from  escaping  laterally. 

But  by  far  the  most  usual  way  of  driving  piles  is  where 
Long  •.  J  r,  I 

pi  les.  they  resist  the  load  by  means  of  the  friction  of  their  sides 
against  the  ground.  In  such  cases  it  is  usual  to  drive  experimental 
piles,  to  ascertain  just  how  much  the  pile  descends  at  the  last  blow  of 
the  hammer  or  ram ;  also  the  amount  of  fall  and  weight  of  ram,  and 


riLKS.  93 

then  to  compute  the  load  the  pile  is  capable  of  resisting :  one-tenth  of 
this  might  be  considered  safe. 
The  formula  tlien  is :  — 

«'  =  nrf  W 

"Where  w  =  the  safe  load  on  each  pile,  in  lbs. 
"        r  =  the  weight  of  ram  used,  in  lbs. 
"        f=  the  distance  the  ram  falls,  in  inches. 
«        s  =  the  set,  in  inches,  or  distance  the  pile  is  driven  at  the 
last  blow. 

"VVliorc  there  is  the  least  doubt  about  the  stability  of  the  pile,  use 
three-fourths  w,  and  if  the  piles  drive  very  unevenly,  use  only  one- 
half  w. 

Some  engineers  prefer  to  assume  a  fixed  rule  for  all  piles.  Profes- 
sor Rankine  allows  200  lbs.  per  square  inch  of  area  of  head  of  pile. 
French  engineers  allow  a  pile  to  carry  50000  lbs.,  provided  it  does 
not  sink  perceptibly  under  a  ram  falling  4'  and  weighing  1350  lbs., 
or  does  not  sink  half  an  inch  under  tliirty  blows.  There  are  many 
other  such  rules,  but  the  writer  would  recommend  the  use  of  the 
above  formula,  as  it  is  based  on  each  individual  experiment,  and  is 
therefore  manifestly  safer. 

Example. 

An  experimental  pile  is  found  to  sink  one-half  inch  under  the  last 
blow  of  a  ram  weighing  1500  lbs.,  and  falling  12'.  What  will  each  pile 
safely  carry  ? 

According  to  formula  (46),  the  safe  load  w  would  be 

w  =  r^r-we  have, 
10.  s 

r==  1500  lbs. 

/=  12.12  =  144",  and 

s  =  ^",  therefore 

1500.144      .„„^^,, 
It;  =  -j^ — =43200  lbs.  ^ 

If  several  other  piles  should  give  about  the  same  result,  we  would 
take  the  average  of  all,  or  else  allow  say  20  tons  on  each  pile.  If, 
however,  some  piles  were  found  to  sink  considerably  more  than 
others,  it  would  be  better  to  allow  but  10  tons  or  15  tons,  according 
to  the  amount  of  irregularity  of  the  soil. 

All  cases  of  pile-driving  require  experience,  judgment,  and  more 
or  less  experiment ;  in  fact  all  foundations  do. 

All  piles  should  be  straight,  solid  timbers,  free  from  projecting 


94  SAFE    BUILDING. 

branches  or  large  knots.     They  can  be  of  hemlock,  spruce  or  wlute 
pine,  but  prcferal)ly,  of  course,  of  yellow  pine  or  oak. 

There  is  danger,  where  they  are  near  the  seashore,  of  their  being 
destroyed  by  worms.  To  guard  against  this,  the  bark  is  sometimes 
shrunken  on ;  that  is,  the  tree  is  girdled  (the  bark  cut  all  around 
near  the  root)  before  the  tree  is  felled,  and  the  sap  ceasing  to  flow, 
the  bark  shrinks  on  very  tightly. 

Others  prefer  piles  without  bark,  and  char  the  piles,  coat  them 
with  asphalt,  or  fill  the  pores  with  creosote.  Cojiper  sheets  arc  the 
best  (and  the  most  expensive)  covering. 

Piles  should  be  of  sufficient  size  not  to  break  in  driving,  and 
should,  as  a  rule,  be  about  30'  long,  and  say  15"  to  18"  diameter  at 
the  top.  They  should  not  be  driven  closer  than  about  2'  6"  in  the 
clear,  or  they  will  be  apt  to  break  the  ground  all  up.  The  feet 
should  be  shod  with  wrought-iron  shoes,  pointed,  and  the  heads  pro- 
tected with  wrought-iron  bands,  to  keep  them  from  sjilitting  under 
the  blows  of  the  ram. 
Sheath-  ^^  sheath-piling  it  is  usual  to  take  boards  (hemlock, 

piling,  spruce,  white  pine,  yellow  pine  or  oak)  from  2"  to  6"  thick. 
Guide-piles  are  driven  and  cross  pieces  bolted  to  the  insides  of  them. 
The  intermediate  piles  are  then  driven  between  the  guide  piles,  mak- 
ing a  solid  wooden  wall  each  side,  from  2"  to  G"  thick.  Sometimes  the 
sheath-piles  arc  tongued  and  grooved.  The  feet  of  the  piles  are  cut 
to  a  point,  so  as  to  drive  more  easily.  The  tops  are  covered  with 
wrought-iron  caps,  which  slip  over  them  and  are  removed  after  the 
piles  are  driven. 

Piles  are  sometimes  made  of  iron :  cast-iron  beincr  T)ref- 
Iron  ^  '  ^  i 

piles,  erable,  as  it  will  stand  longer  under  water.  Screw-piles 
are  made  of  iron,  with  lai'ge,  screw-shaped  flanges  attached  to  the 
foot,  and  they  are  screwed  down  into  the  ground  like  a  gimlet. 

Sheath-piling  is  sometimes  made  of  cast-iron  plates  with  vertical 
strengthening  ribs. 

Where  piles  are  driven  iinder  water,  great  care  must  be  taken 

that  they  are  entirehi  immersed,  and  at  all  times  so.     They  should  be 

cut  off  to  a  imiform  level,  below  the  lowest  low-water  mark.     If  they 

are  alternately  wet  and  dry,  they  will  soon  be  destroyed  by  decay. 

_  After  the  piles  are  cut  to  a  level,  tenons  are  often  cut 

Base-courses  ^  *  ' 

over  piles,  on  their  tops,  and  these  are  made  to  fit  mortises  in  heavy 

wooden  girders  Avhich  go  over  them,  and  on  which  the  superstructure 

The  sizes  given  on  this  page  are  for  heavy  huildiiigs  ;  for  very  light  work  use 
piles  of  8"  to  9"  diameter  at  tops,  about  20  feet  long  aud  16"  apart. 


SLII'-.IOINTS.  95 

rests.  This  is  usiuil  for  docks,  ferry -houses,  etc.  For  other  buildings 
we  fretjuently  see  concrete  packed  hetwecn  and  over  their  tops;  this, 
however,  is  a  very  bad  practice,  as  the  concrete  surrounding  the  tops 
is  apt  to  decay  tliem.  It  is  better  to  cover  the  ])iles  with  3"  x  12"  or 
similar  jjlanks  (well  lag-screwed  to  piles,  where  it  is  necessary  to  steady 
the  latter)  and  then  to  build  the  concrete  base  course  on  these  planks.^ 

Better  yet,  and  the  best  method,  is  to  get  large-sized  building- 
stones,  with  levelled  beds,  and  to  rest  these  directly  on  the  piles.  In 
this  case  care  must  be  taken  that  piles  come  at  least  under  each  cor- 
ner of  the  stone,  or  oftener,  to  keep  it  from  tipping,  and  that  the 
stone  has  a  full  bearing  on  each  pile-head.  On  top  of  stone  build 
the  usual  base-courses. 

Piles  should  be  as  nearly  uniform  as  possible  (particularly  in  the 
case  of  short  piles  resting  on  hard  ground),  for  otherwise  their  re- 
spective powers  of  resistance  will  vary  very  much. 

It  is  well  to  connect  all  very  heavy  parts  of  buildings 
joints,    (such  as  towers,  chimneys,  etc.)  by  vertical  slip-joints 
with  rest  of  building.   The  slip-joint  should  be  carried  through  the 
foundation-walls  and  base-courses,  as  well  as  above. 

Where  there  are  very  high  chimneys  or  towers,  or  unbraced  walls, 
the  foundation  must  be  spread  sufficiently  to  overcome  the  leverage 
produced  by  wind.  These  points  will  be  more  fully  explained  in  the 
chapter  on  "  Walls  and  Piers." 

All  base-courses  should  be  carried  low  enough  to  be 
frost,  below  frost,  which  will  penetrate  from  three  to  five  feet 
deep  in  our  latitudes.  The  reason  of  this  is  that  the  frost  tends  to 
swell  or  expand  the  ground  (on  account  of  its  dampness)  in  all  direc- 
tions, and  does  it  with  so  much  force  that  it  would  be  apt  to  lift  the 
base-course  bodily,  causing  cracks  and  possible  failure  above. 

1  The  strength  of  planks  is  calculated  the  same  as  for  beams  laid  on  their  fiat 
sides. 


96 


SAFE    IIUILUIXG. 


CHAPTER   III. 


CELLAR   AND   RETAINING   WALLS. 


PCB 


f 


Fig.  49. 


ANALYTICAL   SOLUTION. 

rHE  architect  is  sometimes  called  upon  to 
build  retaining-waUs  in  connection  with  ter- 
races, ornamental  bridges,  city  reservoirs, 
or  similar  problems.  Then,  too,  all  cellar  walls, 
where  not  adjoining  other  buildings,  become  re- 
taining walls  ;  hence  the  necessity  to  know  how 
to  ascertain  their  strength.  Some  writers  dis- 
tinguish between  "  face-walls  "  and  "  retaining- 
waUs  " ;  a  face-wall  being  built  in  front  of  and 
against  ground  which  has  not  been  disturbed 
and  is  not  likely  to  slide  ;  a  retaining-wall  being 
a  waU  that  has  a  filled-in  backing.  On  this  the- 
ory a  face-wall  would  have  a  purely  ornamental 
duty,  and  would  receive  no  thrust,  care  being 


taken  during  excavation  and  building-operations  not  to  allow  damp 
or  frost  to  get  into  the  ground  so  as  to  prevent  its  rotting  or  losing  its 
natural  tenacity,  and  to  drain  off  all  surface  or  underground  water. 
It  seems  to  the  writer,  however,  that  the  only  walls  that  can  safely 
be  considered  as  "  face-walls  "  are  those  built  against  rock,  and  that 
all  walls  built  against  other  banks  should  be  calculated  as  retaining- 

walls. 

Most  Economi-  The  cross-section  of  retaining-walls  vary,  accord- 
cal  Section,  ing  to  circumstances,  but  the  outside  surface  of  wall 
is  generally  built  with  a  "  batter  "  (slope)  toicards  the  earth.  The 
most  economical  wall  is  one  where  both  the  outside  and  back  sur- 
faces batter  towards  the  earth.  As  one  or  both  surfaces  become  near- 
ly vertical  the  wall  requu-es  more  material  to  do  the  same  work,  and 


KETAINING   WALLS. 


07 


the  most  extravagant  design  of  all  is  where  the  back  face  batters 
away  from  the  earth ;  of  course,  the  outside  exposed  surface  of  wall 
must  eitlicr  batter  towards  the  earth  (A  B  in  Figure  4D)  or  be 
vertical,  (A  C)  ;  it  cannot  batter  away  from  the  ground,  otherwise 
the  wall  would  overhang  (as  shown  at  A  D).  "Wliere  the  courses  of 
masonry  are  built  at  right  angles  to  the  outside  surface  the  wall  will 
be  stronger  than  where  they  are  all  horizontal. 

Thus,  for  the  same  amount  of  material  in  a  wall,  and  same  height, 
Figure  50  will  do  the  most  work,  or  be  the  strongest  retaining-wall, 
Figure  51  the  next  strongest,  Figure  62  the  next,  Figure  53  next, 

w 


Fig.  51. 


Fig.  52. 


Figure  54  next,  and  Figure  55  the  weakest.    In  Figure  50  and  Fig- 
ure 52  the  ioints  are  at  right  angles  to  the  outside  surface;  in  the 


Fig.  55. 

other  figures  they  are  all  horizontal.     For  reservoirs,  however,  the 
shapes  of  Figures  53  or  55  are  often  employed. 


08 


SAFE   BUILDING. 


To  calculate  the  resistance  of  a  retaining-wall  proceed  as  follows  ' 
Height   of  Line     The 
of  Pressure,     c  e  n  - 

tral  line  ov  axis  of  the 
pressure  O  P  or  p  of 
backing  will  be  at  one- 
third  of  the  height  of 
back  surface,  meas- 
ured from  the  ground 
lines,!  tliat  is  at  O  in 
Figure  5G,  where  A  O 
=  i  AB. 

The  direction  of  the 
pressure-line  (except 
for  reservoirs)  is  usu- 
ally assumed  to  form 
an  angle  of  57°  with 
the  back  surface  of 
wall,  or 

L  POB  =  570. 

For  water  it  is  as- 
sumed normal,  that  is, 
at  right  angles  to  the 
back  surface  of  wall. 

If  it  is  desired,  how-  F'g-  56. 

ever,  to  be  very  exact,  erect  O  E  perpendicular  to  back  surface,  aud 
make  angle  E  O  P,  or  (Z.  x)  =  the  angle  of  friction  of  the  filling-in 
or  backing.     This  angle  can  be  found  from  Table  X. 
Amount  of  pres-      The  amount    (p)  of  the   pressure   P  O  is  found 
-General  £j,qjjj  ^j^g  followino;  formulae : 


sure- 
case. 


If  the  backing  is  filled  in  higher  that  the  wall,'' 


Bacl<ing  higher 
than  Wall. 


.L2 


p=- 


ill-x) 


2      sin^.  y.  sin.  (jj-\-x) 


If  the  backing  is  filled  in  only  to  the  top  level  of  wall, 
lo.U^         sin.  X 


Backing  level 
with  Wall 


(47) 


(48) 


2     *  sin  (?/-|-2  x)* 
{ t/cot.  x-cot  (?/  -\-  2x)  —  i/cot.  y  -  cot  (?/  -|-  2x)  j 

I  WTiere  the  earth  in  front  of  the  outside  surface  of  ^7all  C  D  is  not  packed 
very  solidly  below  the  grade  line  and  against  the  wall,  the  total  height  of  wall 
N  B  (including  part  underground)  should  be  taken,  in  place  of  A  B  (the  height 
above  grade  line). 

-  The  top  slope  of  backing  in  this  case  should  never  form  an  angle  with  the 
horizon,  greater  than  the  friction  angle. 


FUICTIOX    01"    GUOUXD. 


99 


Where  p  =  tho  total  amount  of  pressure,  in  pounds,  per  each  run- 
ning foot  in  length  of  wall. 

Where  tv=  the  -weight,  in  pounds,  per  cubic  foot  of  backing. 

Where  L=the  height  of  retaining   wall  above  ground,  in  feet. 
Sec  foot  note  1,  p.  98. 

AYhere  y  =  the  angle  formed  by  the  back  surface  of  wall  with  the 
horizon. 

Where  x  =  the  angle  of  friction  of  the  backing  as  per  Table  X. 
TABLE  X.i 


Material. 


AVERAGE  (except  water). 

A'ery  compact  earth 

Dry  clay 

Sharp  pebbles 

Dry  loam 

Sharp  broken  stones 

Dry  rammed  earth 

Dry  sand 

Dry  gravel 

Wet  rammed  earth 

Wet  sand 

Wet  gravel 

Roimd  pebbles 

Wet  loam 

Wet  clay 

Salt  water 

llaiu  water 


Weight  per  cubic 
foot. 

Angle  of  friction. 

X. 

w. 

120 

33= 

115 

65" 

100 

45° 

110 

45=> 

100 

40° 

100 

38° 

110 

370 

112 

32° 

110 

32° 

125 

27° 

125 

24° 

125 

24° 

110 

23° 

130 

17° 

125 

17° 

G4 

0° 

C2J 

0=" 

Even  those  who  do  not  understand  trigonometry  can  use  the  above 
formulae. 

It  will  simply  be  necessary  to  add  or  subtract,  etc.,  the  numbers 
of  degrees  of  the  angles  y  and  x,  and  then  find  from  any  table  of 
natural  sines,  cosines,  etc.,  the  corresponding  value  for  the  amount 
of  the  new  angle.  The  value,  so  found,  can  then  be  squared,  multi- 
plied, square  root  extracted,  etc.,  same  as  any  other  arithmetical 
problem.  Should  the  number  of  degrees  of  the  new  angle  be  more 
than  90°,  subtract  90°  from  the  angle  and  use  the  positive  cosine  of 
the  difference  in  place  of  the  sine  of  whole,  or  the  tangent  of  the 
difference  in  place  of  the  co-tangent  of  the  whole ;  in  the  latter  case 
the  value  of  the  tangent  will  be  a  negative  one,  and  should  have  the 
negative  sign  prefixed. 

Thus,  if  a:  =  33°  and  7/ =  50°,  formula  (47)  would  become: 
_  tc.  L2    sin.2  (1 7)° 


p: 


sin.2  50.0  giu,  88° 


'Ab  )ve  table  of  friction  angles  is  taken  from  Klasen's  "  Jlockbau  und  Briick- 
eiibau  -  Conslructlonen."  As  a  rule  it  will  do  to  assume  the  angle  of  friction  at33** 
and  tlie  weight  of  backing  at  120  lbs.  per  cubic  foot,  except  in  the  case  of  water. 


100  SAFE    BUILDING. 

The  values  of  wliicli,  found  in  a  table  of  natural  sines,  etc.,  is  • 
_  u:  U-     0,2924-^ 
P~     2      '    0,7G(i-^.  0,999-4 

■1l.  0,1458      =  0,729.  ?o.L2 


2 

Similarly,  in  formula  (48),  we  should  have  for  tlie  quantity  : 
y/cot.a;-cot.  (^+2x-;  =  y/cot.  33°  -  cot.  110° 

=  y/cot.  33°  -  [-tg.  (110"  -  yo-;j 

=  y/cot.  33°  +  tg.  26° 
__y/l,5399  4-0,4877 
=  y/2,0276     =1,424 
Average  Case.      As  already  mentioned,  however,  the  angle  of  fric- 
tion —  (except  for  water  when  it  is=r  0°,  that  is,  normal  to  the  back 
surface  of  wall)  —  is  usually  assumed  at  33° ;  this  would  reduce  above 
formulae  to  a  very  much  more  convenient  form,  viz. : 
For  the  average  angle  of  friction  (33°) 

If  the  backing  is  higher  than  the  wall :  

Backing  higher         ^^_r.L2.   (10 -77.  0,55)^.  Vl44 -}- n^  (49) 

than  Wall.  ^  2         (10  +  ?i.  0,55).     144 

Or,  if  the  backing  is  level  with  top  of  wall: 

Backing    level       „  _  !^^     Vl44-fn2     /  V  A  5i_  "•  M- H  ..._ 
with  Wall.  ''  2     '     9-}-n.l,7      V^    '  5-|-n.0,9 

i/n     _n.0.4-riy  (50) 

V  12       5  4- n.  0,9/ 
Where  p,  to  and  L  same  as  for  formulae  (47)  and  (48). 
"\Miere  ?i  =  amount  of  slope  or  batter  in  inches  (per  foot  height  of 
wall)  of  rear  surface  of  wall. 

Thus,  if  the  rear  surface  sloped  towards  the  backing  three  inches 
(for  each  foot  in  height)  we  should  have  a  positive  quantity,  or 
71  =  -f  3. 
If  the  rear  surface  sloped  aiva^  from  the  backing  three  inches 
(per  foot  of  height),  n  would  become  negative,  or 
n  =  — 3. 
When  the  rear  surface  of  wall  is  vertical,  there  would  be  no  slope, 
and  we  Avould  have 
Cellar  Walls.        n=0. 

The  latter  is  the  case  generally  for  all  cellar  walls,  which  would 
still  further  simplify  the  formula,  or,  for  cellar  loalls  where  weight 
of  soil  or  backing  varies  materially  from  120  pounds  per  cubic  foot. 


CELLAU   WALLS. 


101 


Cellar  Walls-      p  =  w.  JA  O.l^iS.  (51) 

general  case.       ^ 

For  ceUnr  walls,  where  the  wjight  of  soil  or  backing  can  be  safely 

assumed  to  weigh  120  pounds  per  cubic  foot 

Cellar    Walls-     p=lG?i.L^.  (u2) 

usual  case.         ^  •» 

Where  p  =  the  total  amount   of  pressure,  in   pounds,  per   each 

running  foot  in  length  of  -wall. 

Where  ?r=:the  weight,  in  pounds,  per  cubic  foot  of  backing. 

AVhere  L  =  the  height,  in  feet,  of  ground  line  above  cellar  bottom. 

For  different  slopes  of  the  back  surface  of  retaining  walls  (assum- 
ing friction  angle  at  33°)  we  should  have  the  following  table ;  -(- 
denoting  slope  towards  backing,  -  denoting  slope  away  from  backing. 
TABLE  XI. 


Slope  of   back   sur- 

Value of  p  for  backings  of 

Value  of  p  for  the  average 

face  of  wall  in  inches 

different  weights  per  cubic 

backing,  assumed  to  weigh 

per  foot  of  height. 

foot. 

120  lbs.  per  cubic  foot. 

+4" 

j5=0,072.  w.  Ifl 

V-  H'  L= 

+3" 

p— 0,088.  IV.  \? 

p—W  ,  U 

+2" 

»=0,098.  xo.  L2 

P=i2 .  y 

+1" 

»=0,112.  w.  L2 

;;=13i.  L2 

0" 

j9=0,133.  «•.  L2 

w=163.  L2 

—1" 

;7=0,157.  w.  \? 

p^lQ.  L« 

orr 

p=0,185.  XL\  \? 

n=:22i.  1.2 

3^/ 

^)=0,205.  XL\  \? 

p=2ii.  y 

—i" 

p=0,258.  M?.  L3 

p=31  .  Ifl 

Now  having  found  the  amount  of  pressure  p  from  the  most  conven 
ient  formula,  or  from  Table  XI,  and  referring  back  to  Figure  56, 
proceed  as  follows : 

To  f i  nd  C  u  rve  of  ^'^^^  *^^^  centre  of  gravity  G  of  the  mass  ABC  D,i 
Pressure.  from  G  draw  the  vertical  axis  G  H,  continue  P  O 
till  it  intersects  G  H  at  F.  ]\Iake  F  H  equal  to  the  Aveight  in  pounds 
of  the  mass  A  B  C  D  (one  foot  thick),  at  any  convenient  scale,  and 
at  same  scale  malce  II  I  =  p  and  paral- 
lel to  P  O,  then  draw  I  F  and  it  is  the  I  '^c  ■ 
resultant  of  the  pressure  of  the  earth, 
and  the  resistance  of  the  retaining  walk 
Its  point  of  intersection  K  with  the 
Amount  of         base  D  A  is  a  point  of 

Stressat  Joint,  the  cui'vc   of  pressure,  j^cl— 
To  find  the  exact  amount  of   pressure 
on  the  joint  D  A  use  formula  (44)  for  the 
edge  of  joint  nearest  to  the  point  K  or  edge  D,  and  formula  (45)  for 

the  edge  of  joint  farthest  from  the  point  K,  or  edge  A. 

1  To  lind  the  centre  of  gravity  of  a  trapezoid  A  B  C  D,  Fig.  57,  prolong  CB  until 
B  F  =  CI  =  D  A  and  prolong  D  A  until  A  E  =  D  H  =  C  B.draw  F  I  and  U  Fa:id 
their  point  of  intersection  G  is  the  centre  of  gravity  of  the  whole. 


102 


8AFK   BUILDING. 


•^     a.d 


Formula  (44)  was 
v=JL 

a 

If  ]\I  be  the  centre  of  D  A,  that  is  D  M=  M  A  =  |.  D  A,  and  re- 
membering that  the  piece  of  wall  we  are  calculating,  is  only  one  run- 
ning foot  (or  one  foot  thick),  we  should  have 

For  a;  =  K  INI ;  exjDressed  in  inches. 

For  a  =  A  D.  12 ;  (AD  exjjressed  in  inches). 

For  (/  =  A  D ;  in  inches,  and 

For  jo  =  F  I,  in  lbs.,  measured  at  same  scale  as  F  H  and  H  I ;  or, 
the  stress  at  D,  (the  nearer  edge  of  joint)  would  be  v,  in  pounds,  per 
square  inch, 

AI>.  12~      12.ADa 
Remembering  to  measure  all  parts  in  inches  except  F  I,  which  must 

be  measured  at  same  scale 
G      B^/'"^'^  ^^  '^^'^  "sd  to  lay  out  F  II 

and  II  I.  Similarly  we  should 
obtain  the  stress  at  A  in 
pounds  per  square  inch. 


FI 


6. 


KM.Fl 


AD.  12  "12.AD^ 
V  should  not  exceed  the  safe 
crushing  strength  of  the  ma- 
terial if  positive ;  or  if  y  is 
negative,  the  safe  tensile 
strength  of  the  mortar.  If 
we  find  the  wall  too  weak, 
we  must  enlarge  A  D,  or  if 
too  strong,  we  can  diminish 
it ;  in  either  case,  finding 
the  new  centre  of  gravity  G 
of  the  new  mass  A  B  C  D 
and  repeating  the  operation 
from  that  jioint ;  the  pres- 
sure, of  course,  remaining 
Fig.  58.  the  same  so  long  as  the  sloiie 

of  back  surface  remains  unaltered.  If  the  wall  is  a  very  high  one,  it 
should  be  divided  into  several  sections  in  height,  and  each  section  ex- 
amined separately,  the  base  of  each  section  being  treated  the  same  as 
if  it  were  the  joint  at  the  ground  line,  and  the  ichole  mass  of  wall  bi 
the  section  and  above  the  section  bcinir  taken  in  each  time. 


RESEKVOIR   WALLS.  103 

Thus  in  Figure  58,  when  examining  the  part  A,  BCD,  we  t^houM 
find  O,  P,  for  the  part  only,  using  L,  as  its  height,  the  point  O,  being 
at  one  third  tlie  heiglit  of  A,  B  or  A,  O,  =  ^.  A,  B ;  G,  would  be  the 
centre  of  gravity  of  A,  B  C  D„  while  F,  II,  would  be  equal  to  the 
weight  of  its  mass,  one  foot  thick ;  this  gives  one  point  of  the  curve  of 
pressure  at  K„  with  the  amount  of  pressure  =  F,  I„  so  that  we  can 
examine  the  pressures  on  the  fibres  at  D,  and  A,.  Similarly  when  com- 
paring the  section  A„  B  C  D„  we  have  the  height  L,„  and  so  find  the 
amount  of  pressure  0„  P,„  applied  at  0,„  where  A„  0„  :=  ^  A„  B ; 
G„  is  centre  of  gravity  of  A„  B  C  D„  while  F„  H„  is  equal  to  the 
weight  of  A„  B  C  D,„  one  foot  thick,  and  F„  I„  gives  us  the  amount 
of  pressure  on  the  joint,  and  another  point  K„  of  curve  of  pressure, 
so  that  we  can  examine  the  stress  on  the  fibres  at  D„  and  A„.  For 
the  whole  mass  A  B  C  D  we,  of  course,  proceed  as  before. 
Reservoir  For  reservoirs  the  line  of  pressure  O  P  is  always 

Walls,      at  right  angles  to  the  back  surface  of  the  wall,  so 
that  we  can  simpUfy  formula  (50)  and  use  for  rain  water : 

;j  =  31f     L2  (53) 

For  salt  water : 

;;  =  32.     LS  (54) 

"Where  p  =  the  amount  of  pressure,  in  pounds,  on  one  running  foot 
in  length  of  wall,  and  at  one-third  the  height  of  water,  measured  from 
the  bottom,  and  p  taken  normal  to  back  surface  of  wall, 

"Where  L  =  the  depth  of  water  in  feet. 
If  Backing  Is  "Where  there  is  a  superimposed  weight  on  the  back- 

Loaded  *  ing  of  a  retaining-wall  proceed  as  follows  : 

In  Figure  59  draw 


Si^liil^^»giSf^l^:^ 


&ii^ 


the  angle  C  A  D  =  x, 
the  angle  of  friction  of 
the  material.  Then 
take  the  amount  of 
load,  in  pounds,  com- 
ing on  B  C  and  one 

running  foot  of  it  in   

thickness     (at     right 

angles  to  B  C),  divide 

this  by  the  area,  in  feet,  of  the  triangle  ABC  and  add  the  quotient 

to  w,  the  weight  of  the  backing  per  cubic  foot,  then  proceed  as  before 

inserting  the  sum  w,  in  place  of  w  in  formulaa  (47)  to  (51)  and  in 

2  ~ 
Table  XT,  when  calculating  p ;   or  u\  =  w  -}-  =^^  (55) 

15. Li 


A  D" 

Fig.  59. 


104 


SAFE   BUILDING. 


"Where  w,  =  the  amount,  in  pounds,  to  be  used  in  all  the  formulae 
(47)  to  (51)  and  in  Table  XI,  in  place  of  w. 

"Where  i«  = -weight  of  soil  or  backing,  in  pounds,  per  cubic  foot. 

"\Miere  z  =  the  total  superimposed  load  on  backing,  in  pounds,  per 
running  foot  in  length  of  wall. 

"Where  B  =  length,  in  feet,  of  B  C,  as  found  in  Figure  59. 

"Where  L  =  the  height  of  wall,  in  feet. 

"Where  there  is  a  superimposed  load  on  the  backing,  the  central 
line  of  pressure  j5  should  be  assumed  as  striking  the  back  surface  of 
wall  higher  than  one-third  its  height,  the  point  selected,  being  at  a 
height  X  from  base ;  where  X  is  found  as  per  formula  (56) 
^^L.(e..-§z.) 

2w^-W  ^      ^ 

"Wliere  X  =  the  height,  in  feet,  from  base,  at  which  pressure  is  ap- 
pUed,  when  there  is  a  superimposed  load  on  the  backing. 

"Where  L  =:  the  height,  in  feet,  of  wall. 

"Where  w  =  the  weight,  in  pounds,  per  cubic  foot  of  backing. 

"Where  w,:='is  found  from  formula  (55) 

"When  calculating  the  pressure  against  cellar  walls,  only  the  actual 
weight  of  the  material  of  walls,  floors  and  roof  should  be  assumed  as 
coming  on  the  wall,  and  no  addition  should  be  made  for  wind  nor  for 
load  on  floors,  as  these  cannot  always  be  reUed  on  to  be  on  hand.  The 
additional   compression  due   to  them    should,   however,   be   added 

afterwards.^ 


OBAFHICAL  METHOD. 


X 


^  ^  ^  The  graph- 

^\  ical  method  of 

\      calculating  re- 
\    taining-walls  is 
__A-niuch  easier 
— — """    x''    Lthan  the  ana- 
^^  lytical,  b  e  i  n  g 

less  liable  to 
cause  errors, 
and  is  recom- 
mended  for 
office  use, 
though  the  an- 
alytical method 


1  Where  a  wall  13  not  to  be  kept  braced  until  the  superimposed  wall,  etc.  is  on 
it,  these  should  of  course  be  entirely  omitted  from  the  calculation,  and  the  wall 
must  be  made  heavy  enough  to  stand  alone. 


GRAPHICAL    METHOD. 


105 


might  often  serve  as  a  check  for  detecting  errors,  when  undertaking 
important  work. 

If  A  B  C  D  is  the  section  of  a  retaining  wall  and  13  I  the  top  line 
of  backing,  draw  angle  F  A  M  =  x  =  the  angle  of  friction,  usually 
assumed  at  33°  (except  for  water) ;  continue  B  I  to  its  intersection 
at  E  with  A  M  ;  over  B  E  draw  a  semi-circle,  with  B  E  as  diameter ; 
make  angle  B  A  G  =  2  x  (usually  6G°),  continuing  line  A  G  till  it 
intersects  the  continuation  of  B  I  at  G ;  draw  G  II  tangent  to  semi- 
circle over  BE;  make  G  I  =  G  II ;  draw  I  A,  also  I  J  parallel  to 
B  A  ;  draw  J  K  at  right  angles  to  I  A ;  also  B  ]\I  at  right  angles  to 
A  E.  Now  for  the  sake  of  clearness  we  will  make  a  new  drawing  of 
the  wall  A  B  C  D  in  Figure  61, 

Calling  B  M  =  Z  and  K  J  =  Y  (both  in  Figure  60)  make  A  E  =  Q 
(Figure  61)  where  Q  is  found  from  formula  (57)  followino-:  — 

■\r    r/  \       y  a 

1.  Z.S 


Q  = 

Where  Q  =  the  length  of  A  E  in  Figure  61,  in  feet, 
"Where  Y  =  the  length  of  K  J  in  Figure  60,  in  feet,i 
Where  Z  =  the  length  of  B  M  in  Figure  60,  in  feet, 

C 


(57) 


Figs.  61    and  62. 

Where  s  =  the  weight  of  one  cubic  foot  of  backing,  in  lbs. 
^Vhere  m  =  the  weight  of  one  cubic  foot  of  wall,  in  lbs. 
Where  L  =  the  height  of  backing,  in  feet,  at  walk 


JfR  fw  ^^d  «  f  I  ?-®  ^  }  ^?  ^^"^  lionzon  13  equal  to  the  augle  of  friction,  as  is 
often  the  case,  faml  A  G  as  before  ami  use  tliis  length  in  place  of  K  J  or  Y  which 
of  course,  it  will  bo  impossible  to  flnd,  as  A  M  and  B  I  would  bo  oaral'lel  and 
would  have  no  point  of  intersection  ;  of  course,  B  1  should  never  be  steeper  than 
A  INI  or  else  all  of  the  soil  steeper  than  the  lino  of  angle  of  friction  would  be  apt 


106  SAFE    BUILDING. 

Sections  must  Draw  E  B,  divide  the  wall  into  any  number  of  sec- 
helg^hts.^  tions  of  equal  height,  in  this  case  we  will  say  three 
sections,  A  A.  D,  D ;  A,  A„  D„  D,  and  A„  B  C  D„.  Find  the  cen- 
tres  of  gi-avity  of  the  different  parts,  viz. :  G,  G,  and  G,„  also  F,  F, 
and  F„.  Bisect  D  D,  at  S,  also  D.  D„  at  S,  and  D„  C  at  S„-  Draw 
S  N,  S,  N,  and  S„  N„  horizontally.  Through  G,  G.  and  G„  draw 
vertical  axes,  and  through  F,  F,  and  F„  horizontal  axes,  till  they  in- 
tersect A  B  at  O,  O,  and  0„.  Draw  O  P,  O.  P,  and  0„  P„  parallel 
to  M  A,  where  angle  M  A  E  =  x  =  angle  of  friction  of  soil,  or  back- 
ing. In  strain  diagram  Figure  62  make  a  6,  :=  R„  N„ ;  also  6,  rf,  =-: 
R,  N,  and  d,  /  =  R  N.  From  h„  d,  and  /  draw  the  vertical 
lines.  Now  begin  at  a ;  draw  a  b  parallel  to  M  A ;  make  6  c  =  S„  R„ ; 
draw  c  d  parallel  to  INI  A ;  make  d  e  =  S,  R, ;  draw  e  f  parallel 
to  M  A  and  make  f  g  =z  S  R.  Draw  a  c,  a  d,  a  e,  a /and  a  g.  Now 
returning  to  Figure  Gl,  prolong  P„  0„  till  it  intersects  the  vertical 
axis  through  G„  at  H„ ;  draw  II„  H,  parallel  to  «  c  till  it  intersects 
P,  O,  at  11, ;  draw  II,  I.  parallel  to  a  d  till  it  intersects  the  vertical 
axis  through  G,  at  I, ;  draw  I,  II  parallel  to  a  e  till  it  intersects  P  O 
at  H;  draw  H  I  parallel  to  a /till  it  intersects  the  vertical  through 
G  at  I ;  draw  I  K  parallel  to  a  g.  Then  will  points  K,  K,  and  K„  be 
points  of  the  curve  of  pressure.  The  amount  of  pressure  at  K„  will 
be  a  c,  at  K,  it  will  be  a  e,  and  at  K  it  will  be  a  g,  from  which,  of 
course,  the  strains  on  the  edges  D,  D,  and  D„,  also  A,  A,  and  A„  can 
be  calculated  by  formulaj  (44)  and  (45).  To  obtain  scale,  by  which 
Scale  of  strain  ^o  measure  a  c,  a  e  and  a  g,  make  g  h,  Figure  62  at 
diagram.  anj  scale  equal  to  the  weight,  in  pounds,  of  the  part 
of  wall  A  A,  D,  D  one  foot  thick,  draw  h  i  parallel /or,  then  g  i  meas- 
ured at  same  scale  as  g  h,  is  the  amount  of  pressure,  in  pounds,  at  K. 
Similarly  make  e  k  =:  weight  of  centre  part,  and  c  m  =l  weight  of 
upper  part,  draw  k  I  parallel  d  a,  and  m  n  parallel  b  a,  then  is  e  Z  the 
pressure  at  K,  and  c  n  the  pressure  at  K„,  both  measured  at  same 
scale ;  or,  a  still  more  simple  method  would  be  to  take  the  weight  of 
A  A,  D,  D,  in  pounds,  and  one  foot  thick,  and  divide  this  weight  by 
the  length  of  ^  /in  inches;  the  result  being  the  number  of  pounds 
per  inch  to  be  used,  when  measuring  lengths,  etc.,  in  Figure  62.  The 
above  graphical  method  is  very  convenient  for  high  walls,  where  it  is 
desirable  to  examine  many  joints,  but  care  must  be  taken  to  he  sure  to  get 
the  parts  all  of  equal  height,  otherwise,  the  result  would  be  incorrect. 
If  backing  Ji^  <2^se  of  a  superimposed  weight  find  w„  as  di- 

loaded.  rected  in  formula  (55),  make  A  T  at  any  scale  equal 
to  w  and  A  U  =  «;„  draw  T  E  and  parallel  thereto  U  V,  draw  V  B, 


BUTTRESSED   WALLS. 


107 


tf- 


parallel  to  E  B  and  use  V  B,  in  place  of  E  B,  proceeding  otherwise 
as  before.  The  points  O  of  application  of  pressure  P  O,  will  be 
sliglitly  changed,  particularly  in  the  upper  part,  as  they  will  be  hori- 
zontally opposite  the  centres  of  gravity  of  the  enlarged  trapezoids, 
and  in  the  upper  case  this  point  would  be  much  higher,  the  figure 
now  being  a  trapezoid,  instead  of  a  triangle  as  before. 
Buttressed  ^Vliere  a  wall  is  made  very  thin  and  then  buttressed 
walls,  at  intervals,  all  calculations  can  be  made  the  same  as 
for  walls  of  same  thickness  throughout,  but  the  vertical  axis  through 
centre  of  gravity  of  wall  should  be  shifted  so  as  to  pass  through  the 
centre  of  gravity  of  the  whole  mass,  in- 
.-2ft-.ij    o.^  eluding  buttresses ;  and  the  weight  of  thia 

part  of  wall  should  be  increased  proportion- 
ately to  the  amount  of  buttress,  thus :  If  a 
12"  wall  is  buttressed  every  5  feet  (apart) 
with  2'  X  2'  buttresses,  proceed  as  follows : 
Find  the  centre  of  gravity  G  of  the  part 
of  wall  A  B  C  D  (in  plan)  Figure  63,  also 
centre  of  gravity  F  of  part  E  I II  C,  draw 
lines  through  F  and  G  parallel  to  wall. 
Now  make  a  b  parallel  to  wall  and  at  any 
scale  equal  to  weight  or  area  of  A  B  C  D 

jcai«  of  strains  ovd^"^  ^  ';  ^^"^^  *«  that  of  E  I  H  C.     From 
Fig.  63.  any  point  o  draw  the  lines  oa,ob  and  o  c  ; 

now  drawK  L  (anywhere  between  parallel  hues  F  and  G),  but  paral- 
lel to  b  0,  and  from  L  draw  L  M  parallel  to  o  c,  and  from  K  draw  K 
M  parallel  to  a  o,  a  line  through  their  point  of  intersection  M  drawn 
parallel  to  wall  is  the  neutral  axis  of  the  whole  mass.  When  \)  C 
drawing  the  vertical  section  of  wall-part  A  B  C  D,  Figure 
64,  therefore,  instead  of  locating  the  neutral  axis  through 
the  centre  of  wall  it  will  be  as  far  outside  as  M  is  from  B  C, 
in  Figure  63 ;  that  is,  at  G  H,  Figure  64. 

When  considering  the  weight  per  cubic  foot  of  wall,  we  [^ 

add  the  proportionate  share  of  buttress;  now  in  Figure  63 "^ 
there  are  4  cubic  feet  of  buttress  to  every  7  feet  of  wall,    "^'s-  ^•*' 
so  that  we  must  add  to  the  usual  weight  w  per  cubic  foot  of  wall  ^  w, 
or  w.  (1  -|-f) 

To  put  this  in  a  formula. 


5oo  loop      geoo 


W,,  =  W  (1  +  J-) 


(58) 


108 


SAFE   BUILDING. 


"Wliere  w„  =  the  weight  per  cubic  foot,  in  pounds,  to  be  used  for 
buttressed  walls,  after  finding  the  neutral  axis  of  the  whole  mass. 

Whei*e  ?^  =  the  actual  weight,  in  pounds,  per  cubic  foot  of  the 
material. 

■WTiere  A  =  the  area  in  square  feet  of  one  buttress. 

Where  A,=  the  area  in  square  feet  of  wall  from  side  of  one  but- 
tress to  corresponding  side  of  next  buttress. 
Walls  with         Buttresses,  however,  will  not  be  of  very  much  value, 

counterforts,  unless  they  are  placed   quite  close  together.     But- 
tresses on  the  back  surface  of  a  wall  are  of  very  little  value,  unless 


SJ5W^'^' 


thoroughly  bonded  and  anchored  to  walls ;  these  latter 
are  called  counterforts.   It  is  wiser  and  cheaper  in  most 
cases  to  use  the  additional  masonry  in  thickening  out  the 
lower  part  of  wall  its  entii-e  length. 
Resistance  Where  frost  is  to  be  resisted  the  back 

to  frost,  part  of  wall  should  be  sloped,  for  theTS^ 
depth  frost  is  hkely  to  penetrate  (from  3  to  4  feet  in     'p.g.  65. 
our  chmate),  and  finished  smoothly  with  cement,  and  then  asphalted, 
to  allow  the  frozen  earth  to  slide  upwards,  see  Figure  G5. 
Example  I. 
Cellar  wall  to        -^  ^'■'-'o  story  and  attic  frame  house  has  a  12"  brick 
frame  <iyteUins- foundation  wall,  the  distance  from  cellar  bottom  to 
ground  level  being  6  feet.    The  angle  of  friction  of  ground  to  be  assumed 
at  33°  and  the  weight  per  cubic  foot  at  1 20  pounds.     Is  the  wall  safe  f 
The  weight  of  wall  and  superstructure  must, 
of  course,  be  taken  at  its  minimum,  when  cal- 
culating its  resistance  to  the  ground,  we  shall, 
therefore,  examine  one  of  the  sides  on  which 
no  beams  or  rafters  rest.   The  weight  will  con- 
sist, therefore,  only  of  brick  wall  and  frame 
wall  over.     We  examine  only  one  running  foot 
of  wall,  and  have 
8  cubic  feet  of  brick  at  112  lbs.        =    BOG  lbs 
24  feet  (high)  of  frame  wall  at  15  lbs.  =    3G0  " 
Total  weight  resisting  pressure  =  125G  " 
The  pressure  itseK  will  be  according  to  form- 
ula (52) 

P  =  1G^.  L2=1G§.  36 
=  GOO  lbs. 
f^  Now  make  D  O  =  i   D  C.    Make  angle 

r  O  C  =  57° ;  prolong  P  O  till  it  intersects  the 


I     f!    t«    it    *a    Co 

.Jcale  of  Lan^rhsOntMoS 

Fig.  66 


KXAMI'LES.  109 

vertical  neutral  axis  of  wall'  at  F  ;  make  F  II  =  125C  pounds,  at  any 
scale,  draw  H  I  parallel  to  P  O,  and  make  II  I  =p  =  GOO  pounds  at 
same  scale.  Draw  1  F,  then  is  its  point  of  intersection  K  (with  D  A) 
a  point  of  the  curve  of  pressure,  and  F  I  (measured  at  same  scale)  is 
the  amount  of  pressure  p  to  be  used  in  formula;  (44)  and  (45).  By 
careful  drawing  we  will  find  that  K  comes  ^"  beyond  A  (outside  of 
A  D),  or  G^"  from  centre  E  of  A  D.  F  I  we  find  measures  1G60 
units,  therefore  p  =  IGGO  pounds. 

To  find  the  actual  stress  or  resistance  v  of  edge  of  fibres  of  brick- 
work at  A  use  (44),  viz. : 

;)     I    „    x.p 
a  a.  a 

and  asp  =  16G0  and  x  =  E  K  =  Gi"  and  a  =  12.  12  =  144  inches 
and  rf  =  A  D  =  12"  we  have : 

i;=l^J-G.MlI^=  11  i_L37i  =  +  49  pounds 
144  ~        144.12  i~r      2        -T        i 

as  this  is  a  positive  quantity  it  wiU  be  compression. 

The  resistance  of  edge-fibres  at  D  will  be  according  to  formula  (45) 

1G60       p  Gi  IGGO       -,,       „„,  _„  , 

v= 6.  -^ =  lli  —  37*  =  —  26  pounds 

144  144.  12^2  '■ 

as  this  is  a  negative  quantity  D  will  be  subjected  to  tension ;  that  is, 
there  is  a  tendency  for  A  B  C  D  to  tip  over  around  the  point  A,  the 
point  D  tending  to  rise.  The  amount  of  tension  at  D  is  more  than 
ordinary  brickwork  will  safely  stand,  according  to  Table  V,  still,  as 
it  would  only  amount  to  2G  pounds  on  the  extreme  edge-fibres  and 
would  diminish  rapidly  on  the  fibres  nearer  the  centre,  we  can  con- 
sider the  wall  safe,  even  if  of  but  fairly  good  brickwork,  particularly 
as  the  first-story  beams  and  girders  and  the  end  and  possible  cross- 
walls,  will  all  help  to  stiffen  the  wall.  Had  we  taken  a  foot-slice  of 
the  wall  iinder  the  side  carrying  the  beams,  we  should  have  had  an 
additional  amount  of  weight  resisting  the  pressure.  If  the  beams 
were  18  ft.  span,  we  should  have  three  floors  each  9  ft.  long  and  with 
load  weighing,  say,  90  pounds  per  foot ;  to  this  must  be  added  the 
roof,  or  about  13  ft.  k  50  pounds,  the  additional  load  being: 

Floors,  3.  9.  90  =  2430 

Roof,        13.  50=     650 
Total  3080 

Now  make  I  !M  =  3080  pounds  at  same  scale  as  F  H,  etc.,  draw 
M  F  and  its  point  of  intersection  N  with  D  A  would  be  a  point  of 

>  It  should  really  be  the  vertical  neutral  axis  of  the  whole  weight,  which  would 
be  a  trifle  nearer  to  D  C  thau  centre  of  wall. 


110 


SAFE    UUILDING. 


the  curve  of  pressure,  and  F  M,  at  same  scale  the  amount  of  pressure 
on  D  A  for  the  bearing  walls  of  house;  E  N  we  find  measures  2h", 
and  F  M  measures  4600  units  or  pounds.  The  stress  at  A,  then, 
would  be : 


4600   ,    p  2i.460O 
144  ~        144. 


12 


=  -j-  72  pounds. 


and  the  stress  at  D  would  be  : 


_  4600_g 


2h  4000 


=  —  8  pounds. 


144  144. 12 

There  will,  therefore,  be  absolutely  no  doubt  about  the  safety  of 
bearing  walls. 

Example  IT. 
Cellar  wall  A  cellar  wall  A  B  C  D  is  to  be  carried  15  feet  be- 

lofn\^%t>uf\a^ns'low  the  level  of  adjoining  cellar;  for  particular  rea- 
sons the  neighboring  wall  cannot  be  underpinned.  It  is  desirable  not  to 
mate  the  loall  A  B  C  D  over  2'  4"  thick.  Would  this  be  safe  f  The 
soil  is  wet  loayn. 

In  the  first  place,  before 
excavating  we  must  sheath- 
pile  along  line  C  D,  then 
as  we  excavate  we  must 
secure  horizontal  timbers 
along  the  sheath-piUng  and 
brace  these  from  opposite 
side    of    excavation.     The 


sheath-piling  and  horizon- 
tal timbers  must  be  built 
in  and  left  in  wall.  The 
braces  will  have  to  be  built 
around  and  must  not  be 
removed  until  the  whole 
weight  is  on  the  wall. 

The  weight  of  the  wall 
C  G,  per  running  foot  of 
length,  including  floors  and 
roofs,  we  find  to  be  13000 
pounds,  but  to  this  we  must 
add  the  possible  loads  com- 
ing on  floors,  wliich  we  find 
to  be  6000  pounds  addi- 
tional, or  19000  pounds  total,  possible  maximum  load.    This  load  will 


*», 


4a    60  72  84   96 

Jcale  of=  Lenjrhi  Unchu) 

»0         loocq       ^?°^^         ^QP*^^         ■-— , 

JcaTe  of  JtroinJ(lk>J) 
Fig.  67. 


DEEP    CELLAU     WALL.  HI 

be  distributed  over  the  area  of  C  D  E.     Tii  calculating  the  weight  of 
A  B  C  D  resisting  the  pressure,  we  must  take,  of  course,  only  the  min- 
imum weight ;  that  is,  the  actual  weight  of  construction  and  omit  all 
loads  on  floors,  as  these  may  not  always  be  present.     The  weight  of 
walls  and  unloaded  floors  coming  on  A  B  C  D,  and  including  the 
weight  of  A  B  C  D  itself,  we  find  to  be  21500  j)ounds  per  running 
foot.     Now  to  find  the  pressure  p,  proceed  as  follows  :  Make  angle 
E,  D  M=  17°,  the  angle  of  friction  of  wet  loam  (Sec  Table  X),  and 
prolong  D  E,  till  it  intersects  C  E„  at  PI    Now  C  E,  we  find,  measures 
52  feet;  CD  or  L  is  15  feet ;  then,  instead  of  using  w  iu  formula 
(51),  we  must  use  «?,,  as  found  from  formula  (55),  viz. : 
,     2.  19000 
"■  =  "+(7^7^ 
w  for  wet  loam  (Table  X)  is  130  pounds;  therefore, 

ry,=  130  +  ^1^  =  130 +  48,  7  =  179. 
52.    15  ' 

Inserting  this  value  for  lo  in  formula  (51)  we  have  : 

;>  =  179.  152.  0,138  =  5558. 
The  height  X  from  D  at  which  P  O  is  applied  is  found  from  formula 
5G,  and  is  : 

V  _  15.  (179  —  |.  150)  __  15.  (179  — 100)  _ 
2.179  —  150       "~       358  — 150       ~ 
5',697==:5'8i" 

Make  D  0  =  X  =  5'8r ;  draw  P  O  parallel  to  E  D  till  it  inter- 
sects the  vertical  neutral  axis  of  wall  (centre  line)  at  F  ;  make  F  II 
(vertically)  at  any  scale  equal  to  21500,  draw  H  I  parallel  to  P  O 
and  make  II  1=7^  =  5558  pounds  at  same  scale,  draw  I  F,  then  is 
its  point  of  intersection  with  the  prolongation  of  A  D  at  K  a  point  of 
the  curve  of  pressure,  and  F  I  measui-ed  at  same  scale  as  F  H  is  the 
amount  of  pressure  on  joint.  The  distance  K  from  centre  of  joint  N 
we  find  is  14^",  F  I  measures  23800  units  or  pounds;  the  stress  (y) 
at  A,  therefore,  will  be,  formula  (44)  : 

^  _  23800  14^^^3800  _ 

28.12  ~        28.12.28         ~ 
"While  the  stress  at  D  would  be  formula  (45) 

23800        „  14^.23800 
—  ii. —^ =  — 150 


28.12  28.12.28 

Or  the  edge  at  A  would  be  subjected  to  a  compression  of  292  pounds, 
while  the  edge  at  D  would  be  submitted  to  a  tension  of  150  pounds 
per  square  inch,  both  strains  much  beyond  the  safe  limit  of  even  the 
best  masonry.  The  wall  will,  therefore,  have  to  be  thickened  and  a 
new  calculation  made. 


112 


SAFK   BUILDING. 


Examjde  III. 
Wall  to  -^  stage-pit  ZO  feet  deep  is  to  le  enclosed  by  a  stone- 

stage  pit.  icall,  3  feet  thick  at  the  top  and  increasing  4  inches  in 
thickness  for  every  5  feet  of  depth.  The  wall,  etc.,  coming  over  this 
wall  weighs  25000  pounds  per  running  foot,  hut  cannot  he  included  in 
the  calculation,  as  peculiar  circumstances  will  not  allow  hraces  to  he 
kept  against  the  cellar  wall,  until  the  superimposed  loeight  is  on  it.  The 
surrounding  ground  to  be  taken  as  the  average,  that  is,  120  pounds 
weight  per  cubic  foot,  and  with  an  angle  of  friction  of  33°. 

c 


> 


S" 


//A 


rM- 


\;^-P' 


Ff:4. 


^^■^^ih 


a<:. 


/Pu 


\ 


t^. 


:^ 


y^\ 


^r-'-'f 


o  i2  Zi-^6  48  (fa  77  S» 

Jcale  of  Lengfhs  (inches) 
Fig.  68. 

Find  B  M  (=  Z)  and  Q  T  (=  Y)  by  making  angles  T  A  E  =  330 
and  B  A  U  =  66°  and  then  proceeding  as  explained  for  Figure  60. 
We  scale  B  M  and  Q  T  at  same  scale  as  height  of  wall  A  B  is  drawn, 
and  find : 

BM  =  Z  =  25ft.  6"  =  25J 

QT  =  Y  =  9  ft.  8"  =  9|;  assuming  each  cubic  foot  of 
wall  to  weigh  150  pounds  we  find  Q  from  Formula  (57) 


8TAGK-PIT   WALL. 


113 


Q  =  93- 25^.120  _  Qr^  yg  _  g,  y,,^ 

^  30.  150  ' 

Make  A  E  =  Q  =  G'  7"  and  draw  B  E. 

At  equal  heighls,  that  is,  every  5  feet,  in  this  case,  draw  the  joint 
nes  D  E,  D,  E„  D„  E,„  etc.     Find  the  centres  of  gravity  F,  F„  F,„ 

etc.,  of  the  six  parts  of  A  E  B, 
"^'^(see  foot-note,  p.  101)  and  also 
the  centres  of  gravity  G,  G„  G,„ 
etc.,  of  the  six  parts  of  the  wall 
itself,  which,  in  the  latter  case, 
will  be  at  the  centre  of  each 
part. 

Horizontally,  opposite  the 
centres  F,  F„  F,„  etc.,  apply 
the  pressures  P  O,  P,  0„  etc., 
against  wall,  and  parallel  to 
M  A.  Through  centres  G,  G„ 
G,„  etc.,  draw  vertical  axes. 

Draw  the  lines  S  N,  S,  N,, 
S„  N,,,  etc.,  at  half  the  vertical 
height  of  each  section.  Now  in 
Figure  G  9  make  a  ft,  =  Rv  Nv ; 
fe.  d,  =  B.v  N,v ;  (1,  /,  =  R,„  N„, ; 
/, /j,  =  R„N„;  /^y,  =  R.N.; 
and  j\  I,  =  R  N.  Draw  the 
vertical  lines  through  these 
points.  Now  begin  at  a,  make 
a  b  parallel  to  P  O,  make  b  c=i 
Rv  Sv ;  draw  c  d  parallel  P  0  ; 
make  de  =  R,v  S,v ;  draw  e  f 


Jcale  op  Lengths  (Inches) 
Fig.  69. 


parallel  P  0,  make  fgz=  R„,  S„„  and  similarly  gh,i  j  and  k  I  par- 
allel P  O,  and  h  i  =  R„  S,. ;  y  A:  =  R,  S,  and  Z  m  =  R  S.  Draw  from 
a  lines  to  all  the  points  c,  d,  e,  f,  g,  etc.  Now  in  Figure  G8  begin  at 
Pt  Ov,  prolong  it  till  it  intersects  vertical  axis  Gy  at  Iv,  draw  I^  H^ 
parallel  a  c  till  it  intersects  P,y  0,v  at  11,.;  draw  Ily  T,v  parallel  to  a  d 
till  it  intersects  vertical  axis  G,v  at  T,v  ;  draw  1,^  Il.y  parallel  to  a  e 
tm  it  intersects  P,„  0„,  at  H,^  and  similarly  II.v  I,„  parallel  af;  I,„  11,,, 
parallel  ag:  II,„  T„  parallel  a  h;  I„  II„ parallel  to  a  i;  II,.  I,  parallel 
to  a  j ;  1, 11,  parallel  to  a  k  ;  II,  I  parallel  to  a  I,  and  I K  parallel  to  a  m. 
The  points  of  intersection  K,  K„  K,„  etc.,  are  points  of  the  curve 
of  pressure.     To  find  the  amount  of  the  pressure  at  each  point,  find 


114  SAFE    BUILDING. 

weif^ht  per  running  foot  of  length  of  any  part  of  wall,  say,  the  bottom 
part  (A  A,  D,  D)  the  contents  are  5'  high,  4'  8"  wide,  1'  thick  =:  5.4§.  1 
=  23^  cubic  feet  a  150  pounds 
=  3500  pounds. 

Divide  the  weight  by  the  length  of  m  I  in  inches,  and  we  have  the 
number  of  pounds  per  inch,  by  Avhich  to  measure  the  pressures.  As 
m  I  measures  50  inches,  each  inch  will  represent  ^|§^  =  C2|-  lbs. 

Xow  let  us  examine  any  joint,  say,  A,„  D„, ;  I„,  II„,  which  inter- 
sects A„,  D,„  at  K,„  is  parallel  to  a  g.  Now  a  g  scales  166  inches, 
therefore,  pressure  at  K,„=:  1G6.  02^  =  10375  pounds.  In  measur- 
ing the  distance  of  K,„  from  centre  of  joint  in  the  following,  remem- 
ber that  the  width  of  A,„  D„,  is  44  inches,  the  width  of  masonry  above 
joint,  and  not  48"  (the  width  of  masonry  below).  A,„  K,„  scales  38", 
therefore,  distance  x  of  K,„  from  centre  of  joint  is  x  =  38  —  22  :^  16". 

We  have,  then,  from  Formula  (44) 

,-p,  10375,.    16.10375  ,    .„  , 

stress  at  D,„    v ^=  ^  ,  ^  -  +  6.   .  ,  ^  .  ,,    =  -4- 63  pounds. 
44.12  44.12.44  '         ^ 

and  from  formula  (45) 

,   .  10375        „    16.10375  „„  , 

stress  at  A... ;  v  =  ^^^^^  —  0.  -^^j^j^  =  —  23  pounds. 

The  joint  A„,  D„„  therefore,  would  be  more  than  safe. 

Let  us  try  the  bottom  joint  A  D  similarly.     I  K  is  parallel  to  a  m 
now  a  m  scales  480",  therefore,  the  pressure  at  K  is  ^  ^=480.  62^^  =z 
30000  pounds. 

Now  K  is  distant  53  inches  from  centre  of  joint,  therefore,  stress 

•     T^  •  30000    ,    .    53.30000  ,    ^„„  , 

^' ^  ^^  "  =  3gT2- + '^^  5(U23(r  =  + -^^  P°"^^^- 

,    ,  ...  30000        „    53.30000  „^„  , 

and  stress  at  A  is  u  = —  6. =  —  209  pounds. 

oG.12  5G. 12.56  ^ 

The  wall  would  evidently  have  to  be  thickened  at  the  base.     If  we 

could  only  brace  the  wall  until  the  superimposed  weight  were  on  it, 

this  might  not  be  necessary.     If  we  could  do  this  we  should  lengthen 

6  c  an  amount  of  inches  equal  to  the  amount  of  this  load  divided  by 

62J  (the  number  of  pounds  per  inch),  or  6  c  instead  of  being  36  inches 

long  would  be : 

36  +  -r^^  =  436  inches  long. 

While  this  lengthening  of  /;  c  would  make  the  lines  of  pressure  a  c,ae, 
af,  etc.,  very  much  longer,  and  consequently  the  actual  pressure  verv 
much  greater,  it  will  also  make  them  very  much  steeper  and  conse- 
quently bring  this  pressure  so  much  nearer  the  centre  of  each  joint. 


EXAMTLE-KESERVOIR. 


115 


C   5 


that  the  pressure  will  distribute  itself  over  the  joint  mui'h  more 
evenly,  and  the  worst  danger  (from  tension)  -will  probably  be  entirely 
removed. 

Example  IV. 
Reservoir  ^  stone  reservoir  loall  is  plumb  on  the  outside,  2  feet 

Walli    wide  at  the  top  and  5  feet  loide  at  the  bottom;  the  wall 
is  21  feet  high,  and  the  possible  depth  of  loater  20  feet.   Is  the  wall  safe  ? 

Divide  the  wall  into  three  parts  in  height ;  that  is,  D  D,  =  D,  D„  r=: 
D„  C.  Find  the  weight  of  the  parts  from  eaeh  joint  to  top,  per  run- 
ning foot  of  length  of 
wall,  figuring  the  stone- 
work at  150  pounds  per 
cubic  foot,  and  we  have: 

Weight  of  A„  BCD,, 
=  2975  pounds. 

Weight  of  A.  BCD. 
=  7140  pounds. 

Weight  of  A  B  C  D 
=  12495  pounds. 

Find  centres  of  grav- 
ity of  the  parts  A  B  C  D 
(at  G),  of  A,  B  C  D,  (at 
G.)  and  of  A„  B  C  D„ 
(at  G„).  Apply  the 
pressures  P  O  at  ^ 
height  of  A  E ;  P,  O.  at 
^  height  of  A,  E  and 
P„  0„  at  A  height  of 
A„  E,  where  E  top  level 
of  water. 

The  amount  of  pres- 
sures will  be  from  form-  / 
ula  (53).  r"' 

For  part  A„  E. 
P..  0„  =  31|.  L2,  ~ 
31|.  C2=1125  poimds. 


— \^- 

Dii 

binii°Gi  r            , 

1  Xl- 

'  o 

/ 

1  I 

if — 

.__;l--\ai-— ^-ip 

■7%        \ 

1' 

\          \ 

1 

A 

D    in     A 

H 


JcQle  of  itTfiinj  ()lps) 

J<.ale  of-  Lengths  (.m'hfs) 
Fig.  70. 

For  part  A,  E  ;  P.  O,  =  31f  L^,  =  31f  132  _  5251  pounds. 
For  part  A  E  ;  P  O  =  31^.   1.2=  31 1.    202  _  JOSOO  pounds. 
•  The  pressures  P  O,  P,  0„  et-.,  will  be  applied  at  right  angles  to 
A  E,  prolong  these  lines,  till  they  intersect  the  vertical  axes  through 


116  SAFE   BUILDING. 

(the  centres  of  gravity)  G,  G,  and  G„  at  F,  F,  and  F„.     Then  make 

F„  II„  =    2D  75,  weight  of  upper  part. 

F,  II,  =    7140,  weight  of  A,  B  C  D„  and 

F   II   =  12405,  weiglit  of  A  B  C  D. 
Draw  through  II,  II,  and  II„  the  hncs  parallel  to  pressure  lines 
making 

H„I„==P„0„=    1125 

II,  I,  =  P,  O,  =    5281 

II  I    ^P    O   ==12500 
Draw  I„  F,„  I,  F,  and  I  F,  then  will  their  lengths  represent  the 
amounts  of  pi-essure  at  points  K,„  K,  and  K  on  the  joints  A„  D„,  A,  D, 
and  A  D. 

F„  I„  measures    3300  units  or  pounds. 

F,  I.         "  9500     "         " 

F    I  "  18800     "         «« 

By  scaling  we  find  that 

K„  is  10^  inches  from  centre  of  D„  A„ 

K,  is  87"  »  "     D,  A, 

K    is  74  "  "DA 

The  stresses  to  be  exerted  by  the  wall  will,  therefore,  be 

.tD.;.=  -i  +  S^S^  =  +03  pounds. 

atA    ;.  =  i^8^-6.iH8800_  =  _lG7  pounds. 
'  GU.12  60.12.G0  '■ 

From  the  above  it  would  appear  that  none  of  the  joints  are  subject 

to  excessive  compression  :  further  that  joint  D„A„  is  more  than  safe, 

but  that  the  joints  D,  A,  and  D  A  are  subject  to  such  severe  tension 

that  they  cannot  be  passed  as  safe.     The  wall  should,  therefore,  be 

redesigned,  making  the  upper  joint  lighter  and  the  lower  two  joints 

much  wider. 


WALLS  A^D  riEBS.  117 


CHAPTER  IV. 


WALLS  AND   PIERS. 


WALLS  are  usually  built  of  brick  or  stone,  wliicli  are  sometimes, 
tliougli  rarely,  laid  up  dry,  but  usually  with  mortar  filling  all 
the  joints.     The  object  of  mortar  is  threefold  : 
Object  of  1.  To  keep  out  wet  and  changes  of  temperature  by 

mortar,  filling  all  the  crevices  and  joints. 

2.  To  cement  the  whole  into  one  mass,  keeping  the  several  parts 
from  separating,  and, 

3.  To  form  a  sort  of  cushion,  to  distribute  the  crushing  evenly,  tak- 
ing up  any  inequalities  of  the  brick  or  stone,  in  their  beds,  which  might 
fi-acture  each  other  by  bearing  on  one  or  two  spots  only. 

To  attain  the  first  object,  "  grouting  "  is  often  resorted  to.  That  is, 
the  material  is  laid  up  with  the  joints  only  partly  filled,  and  Hquid 
cement-mortar  is  poured  on  till  it  runs  into  and  fills  all  the  joints. 
Theoretically  this  is  often  condemned,  as  it  is  apt  to  lead  to  careless 
and  dirty  work  and  the  overlooking  of  the  filling  of  some  parts ;  but 
practicalUj  it  luakcs  the  best  work  and  is  to  be  recommended,  except, 
of  course,  in  freezing  Aveather,  when  as  little  water  as  possible  should 
be  used. 

To  attain  the  second  object,  of  cementing  the  Avhole  into  one  mass, 
it  is  necessary  that  the  mortar  should  adhere  firmly  to  all  parts,  and 
this  necessitates  soaking  thoroughly  the  bricks  or  stones,  as  other- 
wise they  will  absorb  the  dampness  from  the  mortar,  which  will 
crumble  to  dust  and  fail  to  set  for  want  of  water.  Then,  too,  the 
brick  and  stone  need  washing,  as  any  dust  on  them  is  apt  to  keep  the 
mortar  from  clinching  to  them.  In  winter,  of  course,  all  wetting  must 
be  avoided,  and  as  the  mortar  will  not  set  so  quickly,  a  little  lime  is 
added,  to  keep  it  warm  and  prevent  freezing. 

Thickness  of       To  attain  the  third  object,  the  mortar  joint  must  be 

Joints,  matle  thick  enough  to  take  up  any  inequalities  of  the 

brick  or  stone.     It  is,  therefore,  impossible  to  set  any  stanJard  for 


118  SAFE   BUILDING. 

joints,  as  the  more  irregular  the  beds  of  the  brick  or  stone,  the  hirger 
should  be  the  joint.  For  general  brickwork  it  will  do  to  assume  that 
the  joints  shall  not  average  over  one-quarter  of  an  inch  abwe  irregu- 
larities. Specify,  therefore,  that,  say,  eight  courses  of  brick  laid  up 
*'  in  the  wall "  shall  not  exceed  by  more  than  two  inches  in  height 
eight  courses  of  brick  laid  up  "  dry."  For  front  work  it  is  usual  to 
gauge  the  brick,  to  get  them  of  exactly  even  width,  and  to  lay  them 
up  with  one-eighth  inch  joints,  using,  as  a  rule, "  putty  "  mortar.  While 
this  makes  the  prettiest  wall,  it  is  the  weakest,  as  the  mortar  has  little 
strength,  and  the  joint  being  so  small  it  is  impossible  to  bond  the 
facing  back,  except  every  five  or  six  courses  in  height.  "  Putty  " 
mortar  is  made  of  lime,  water  and  white  Isad,  care  being  taken  to 
avoid  all  sand  or  grit  in  the  mortar  or  on  the  beds. 
Quality  of  "^^^^  ^'^^^  mortar  consists  of  English  Portland  ce- 

mortars.  ment  and  sharp,  clean,  coarse  sand.     The  less  sand 
the  stronger  the  mortar. 

Sand  for  all  mortars  should  be  free  fi'om  earth,  salt,  or  other  impu- 
rities. It  should  be  carefully  screened,  and  for  very  important  work 
should  be  washed.  The  coarser  and  sharper  the  sand  the  better  the 
cement  will  stick  to  it.  English  Portland  cement  will  stand  as  much 
as  three  or  four  parts  of  sand.  Next  to  English  come  the  German 
Portland  cements,  which  are  nearly  as  good.  Then  the  American 
Portland,  and  lastly  the  Rosendale  and  Virginia  cements.  Good 
qualities  of  Piosendale  cements  wiU  stand  as  much  as  two-and-a-half 
of  sand.  Of  lime?,  the  French  lime  of  Tcil  is  the  strongest  and  most 
expensive.  Good,  hard-burned  lime  makes  a  fairly  good  mortar.  It 
should  be  thoroughly  slacked,  as  otherwise,  if  it  should  absorb  any 
dampness  afterwards,  it  will  begin  to  burn  and  swell  again.  At  least 
forty-eight  hours  should  be  allowed  the  lime  for  slacking,  and  it  is 
very  desirable  to  strain  it  to  avoid  unslacked  lumps.  Lime  will  take 
more  sand  than  oement,  and  can  be  mixed  with  from  two  to  four  of 
sand,  much  depending  on  the  quality  of  the  sand,  and  particularly  on 
the  "fatness  "  of  the  lime.  It  is  better  to  use  plenty  of  sand  (with 
lime)  rather  than  too  little ;  it  is  a  matter,  however,  for  practical 
judgment  and  experiment,  and  while  the  specification  should  call  for 
but  two  parts  of  sand  to  one  of  lime,  the  architect  should  feel  at  lib- 
erty to  allow  more  sand  if  thought  desirable.  Lime  and  Rosendale 
cement  are  often  mixed  in  equal  proportions,  and  from  three  to  five 
parts  of  sand  added ;  that  is,  one  of  lime,  one  of  cement,  and  three  to 
five  of  sand.  It  is  advisable  to  specify  that  all  parts  shall  bs  actually 
measured  in  barrel.-:,  to  avoid  such  tricks,  for  instance,  as  hiring  a 


M0RTAK8.  119 

decrepit  laborer  to  sliovcl  cement  or  lime,  while  two  or  tlirec  of  the 
strongest  laborers  are  shovelling  sand,  it  being  called  one  of  cement  to 
two  or  three  of  sand.   A  little  lime  should  be  added,  even  to  the  very 
best  mortars,  in  winter,  to  prevent  their  freezing. 
Frozen  When  a  wall  has  been  frozen,  it  should  be  taken 

walls.  down  and  re-built.  Never  build  on  ice,  but  use  salt, 
if  necessary,  to  thaw  it;  sweep  off  the  salt-water,  which  is  apt  to  rot 
the  mortar,  and  then  take  off  a  few  courses  of  brick  before  continu- 
ing the  work.  Protect  wallf  from  rain  and  frost  in  winter  by  using 
boards  and  tarpaulins.  Some  writers  claim  that  it  does  no  harm  for 
a  wall  to  freeze ;  this  may  be  so,  provided  all  parts  freeze  together 
and  are  kept  frozen  until  set,  and  that  they  do  not  alternately  freeze 
and  thaw,  which  latter  will  undoubtedly  rot  the  inortar. 

Plaster-of- Paris  makes  a  good  mortar,  but  is  expensive  and  cannot 
stand  dampness.  Cements  or  limes  that;  will  set  under  water  are 
called  hydraulic. 

Quickness  of  setting  is  a  very  desirable  point  in  cements.  All  ce- 
ment-mortars, therefore,  must  be  used  perfectly'  fresh;  any  that  has 
begun  to  set,  or  has  frozen,  should  be  condemned,  though  many  con- 
tractors have  a  trick  of  cutting  it  up  and  using  it  over  with  fresh 
mortar.  To  keep  dampness  out  of  cellar-walls  the  outside  should  be 
plastered  with  a  mortar  of  some  good  hydraulic  cement,  with  not 
more  than  one  part  of  sand  to  one  part  of  cement ;  this  cement  should 
be  scratched,  roughened,  and  then  the  cement  covered  outside  with 
a  heavy  coat  of  asphalt,  put  on  hot  and  with  the  trowel.  In  brick 
walls,  the  coat  of  cement  can  be  omitted  and  the  joints  raked  out, 
the  asphalt  being  applied  directly  against  the  brick.  This  asphalt 
should  be  made  to  form  a  tight  joint,  with  the  slate  or  asphalt  damp- 
course,  which  is  built  through  bottom  of  wall,  to  stop  the  rise  of 
dampness  from  capillary  attraction. 

In  ordinary  rubble  stonework  the  mortar  should  be  as  strong  as 
possible,  as  this  class  of  work  depends  entirely  on  the  mortar  for  its 
strength. 

For  the  strengths  of  different  mortars,  see  Table  V. 

Some  cements  are  apt  to  swell  in  setting,  and  should  be  avoided. 

Smoke  Where  flues  or  unplastered  walls  are  built,  the  joints 

flues,  should  be  "■  strucl:"  that  is,  scraped  smooth  with 
the  trowel.  No  flues  should  be  "  pargetted  " ;  that  is,  plastered  over, 
as  the  smoke  rots  the  mortar,  particles  fall,  and  the  soot  accumulating 
in  the  crevices  is  apt  to  set  fire  to  the  chimney.  Joints  of  chimneys 
are  liable  to  be  eaten  out  from  the  same  reason,  and  the  loose  por- 


120  SAFE    BUILDING. 

tions  fall  or  are  scraped  out  when  the  flues  arc  cleaned,  leaving  dan- 
gerous cracks  for  fire  to  escape  through.  It  is  best,  therefore,  to  line 
up  chimneys  inside  with  burned  earthenware  or  fire-clay  pipes.  If 
iron  pipes  are  used,  cast-iron  is  preferable;  wrought-iron,  unless  very 
thick,  will  soon  be  eaten  away.  Where  walls  are  to  be  plastered,  the 
joints  are  left  as  rough  as  possible,  to  form  a  good  clinch. 

^g,,  Outside  walls  are  not  plastered  directly  on  the  in- 

furrings.  side,  unless  hollow;  otherwise,  the  dampness  would 
strike  through  and  the  plaster  not  only  be  constantly  damp,  but  it 
would  ultimately  fall  o£E.  Outside  walls,  unless  hollow,  are  always 
"  furred."  In  fireproof  work,  from  one  to  four-inch  thick  blocks  are 
used  for  this  purpose.  These  blocks  are  sometimes  cast  of  ashes, 
lime,  etc.,  but  are  a  very  poor  lot  and  not  very  lasting.  Generally 
they  are  made  of  burnt  clay,  fire-clay  or  porous  terra-cotta.  The 
latter  is  the  best,  as,  besides  the  advantages  of  being  lighter,  warmer 
and  more  damp-proof,  it  can  be  cut,  sawed,  nailed  into,  etc.,  and  holds 
a  nail  or  screw  as  firmly  as  wood.  These  blocks  are  laid  up  inde- 
pendently of  the  wall,  but  occasionally  anchored  to  the  same  by  iron 
anchors.     The  plastering  is  applied  directly  to  the  blocks. 

In  cheaper  and  non-fireproof  work,  f  urrings  are  made  of  vertical 
strips  of  wood  about  two  inches  wide,  and  from  one  to  two  inches 
tliick,  according  to  the  regularity  of  the  backing.  For  very  fine 
work,  sometimes,  an  independent  four-inch  frame  is  built  inside  of 
the  stone-wall,  and  only  anchored  to  same  occasionally  by  iron  an- 
chors. Where  there  are  inside  blinds,  a  three  or  four  inch  furring  is 
used  (or  a  fireproof  furring),  and  this  is  built  on  the  floor  beams,  as 
far  inside  of  the  wall  as  the  shutter-boxes  demand.  To  the  wooden 
furrings  the  laths  are  nailed.  Furrings  arc  set,  as  a  rule,  sixteen 
inches°apart,  the  lath  being  four  feet  long;  this  affords  four  nailings 
to  each  lath.  Sometimes  the  furrings  are  set  twelve  inches  apart, 
affording  five  nailings.  All  ceilings  arc  cross-furred  every  twelve 
inches,  on  account  of  stiffness,  and  the  strips  should  not  be  less  than 
one-and-thrce-eighths  inches  thick,  to  afford  strength  for  nailing. 
Furring-strips  take  up  considerable  of  the  strain  of  settlements  and 
shrinkage,  and  prevent  cracks  in  plastering  by  distributing  the  strain 
to  several  strips.  To  still  further  help  this  object,  the  -'heading- 
joints  "  of  lath  should  not  all  be  on  the  same  strip,  but  should  bo  fre- 
quently broken  (say,  every  foot  or  two),  and  should  then  be  on  some 
other  strip.  Laths  should  be  separated  sufficiently  (about  three-eighths 
inch)  to  allow  the  plaster  to  be  well  worked  through  the  joint  and  get 
a  strong  grip  or  "  clinch  "  on  the  back  of  the  laths.     If  a  building  is 


SIIUIXKAGE    OP'    WALLS.  121 

properly  built,  llieoretically  correct  in  every  respect,  it  should  not  show 

a  single  crack  in  plastering.     Practically,  however,  this  is  impossible. 

"^  But  there  never  need  be  any  fear  of  shrinkage  or 

Shrinkage  ,,  ',  i     -i  i-  i  ^.^  ^ 

of  joints,  settlement,  in  a  well-constructed  building,  where  tLe 

foundations,  joints  and  timbers   arc   properly  proportioned.     The 
danf^er  is  never  from  the  amount  of  settlements  or  shrinkage,  but  from 
the  °inequnlluj  of  same  in  different  parts  of  the  building.    Inequalities 
in  settlements  are  avoided  by  properly  proportioning  the  foundations. 
Inequalities  in  shrinkage  of  the  joints,  though  quite  as  important,  are 
frequently  overlooked  by  the  careless  architect.    lie  will  build  in  the 
same  budding  one  wall  of  brick  with  many  joints,  another  of  stones 
of  aU  heights  and  with  few  joints,  and  then  put  iron  columns  in  the 
centre,  miking  no  allowance  whatever  for  the  difference  in  shrink- 
age.   If  ho  makes  any,  it  is  probably  to  call  for  the  most  exact  set- 
tincr  of  the  columns,  for  the  hardest  and  quickest-setting  Portland  ce- 
ment for  the  stonework,  and  probably  be  content  with  lime  for  the 
brickwork.     To  avoid  uneven  shrinkages,  allowances  should  be  made 
for  same.     Brickwork  will  shrink,  according  to  its  quality,  from  one- 
sixteenth  to  one-eighth  inch  per  story,  ten  to  twelve  feet  high,  and  ac- 
cording to  the  total  height  of  wall.    The  higher  the  wall,  the  greater 
the  weight  on  the  joints  and  the  greater  the  slirinkagc.     Iron  col- 
mnns  should,  therefore,  be  made  a  trille  shorter  than  the  story  re- 
quires, the  beams  being  set  out  of  level,  lower  at  the  column.     The 
plan  should  provide  for  the  top  of  lowest  column  to  be  one-sixteenth 
or  one-eighth  inch  low,  while  the  top  of  highest  column  would  be  as 
many  times  one-sixteenth  or  one-eighth  inch  low  as  there  were  sto- 
ries ;  or  if  there  were  eight  stories,  the  top  of  bottom  column  for  the 
very  best  brickwork  would  be,  say,  one-sixteenth  inch  low,  and  the 
top  of  highest  column  would  be  one-half  inch  low.   Stone  walls  should 
have  stone  backings  in  courses  as  high  as  front  stones,  if  possible ;  if 
not,  the  backing  should  be  set  in  the  hardest  and  quickest-setting  ce- 
c,i  .  ment.     Stone  walls  should  be  connected   to   brick 

"  joints,  walls  by  means  of  slip-joints.  By  this  method  the 
writer  has  built  a  city  stone-front,  some  150  feet  high  and  over  50 
feet  wide,  connected  to  brick  walls  at  each  side,  without  a  single  stone 
sill,  or  transom,  or  Untel  cracking  in  the  front.  The  slip-joint  should 
carry  through  foundations  and  base  courses  where  the  pressure  is  not 
equal  on  all  parts  of  the  foundation.  If  for  the  sake  of  design,  it  is 
necessary  to  use  long  columns  or  pilasters,  in  connection  with  coursed 
stone  backings,  the  columns  or  pilasters  must  either  be  strong  enough 
to  do  the  whole  work  of  the  wall,  or  else  must  be  bedded  in  putty- 


122  SAFE   BUILDING. 

mortar  with  generous  top  and  bottom  joints,  to  allow  for  shrinkage 
of  the  more  frequent  joints  behind  them ;  otherwise,  they  are  apt  to 
he  shattered.  Such  unconstructional  designs  had,  however,  better  be 
avoided.  In  no  case  should  a  wall  be  built  of  part  iron  uprights  and 
part  masonry;  one  or  the  other  must  be  strong  enough  to  do  the 
work  alone ;  no  rehance  could  be  placed  on  their  acting  together.  In 
Shrinkage  frame  walls,  care  should  be  taken  to  get  the  amount 
of  timber,  of  "  cross  "  timbering  in  inner  and  outer  walls  about 
equal,  and  to  have  as  little  of  it  as  possible.  Timber  will  shrink 
"  across  "  the  grain  from  one-fourth  to  one-half  inch  per  foot.  Where 
the  outer  walls  are  of  masonry,  and  inner  partitions  or  girders  are  of 
wood,  great  care  must  be  taken  that  the  shrinkage  of  each  floor  is 
taken  up  by  itself.  If  the  shrinkage  of  all  beams  and  girders  is  trans- 
ferred to  the  bottom,  it  makes  a  tremendous  strain  on  the  building 
and  will  ruin  the  plastering.  To  effect  this,  posts  and  columns  should 
bear  cUrecthj  on  each  other,  and  the  girders  be  attached  to  their  sides 
or  to  brackets,  but  by  no  means  should  the  girder  run  between  the 
upper  and  lower  posts  or  columns.  If  there  are  stud-partitions,  the 
head  pieces  should  be  as  thin  as  possible,  and  the  studs  to  upper  par- 
titions should  rest  direclly  on  the  head  of  lower  partitions. 
All  beds  ^"^  masonry,  all  beds  should  be  as  nearly  level  as 

level,     possible,  to  avoid  unequal  crushing.     Particularly  is 
this  the  case  with  cut  stonework.     If  the  front  of  the  stone  comes 
closer  than  the  backing  (which  is  foolishly  done  sometimes  to  make  a 
smaU-looking  joint),  the  face  of  the  stone  will  surely  split  off.     If  the 
back  of  a  joint  is  broken  off  carelessly,  and  small  stones  inserted  in 
the  back  of  a  joint  to  form  a  support  to  larger  stones,  they  will  act  as 
wedges,  and  the  stone  will  crack  up  the  centre  of  joint  and  wall. 
Stones  should  be  bedded,  therefore,  perfectly  level  and  solid,  except 
the  front  of  joint  for  about  three-fourth  inches  back  from  the  face, 
which  should  not  be  bedded  solid,  but  with  "  putty  "-mortar.     Light- 
Cement  colored  stones,  particularly  lime-stones,  are  apt  to 
stains,  stain  if  brought  in  connection  with  cement-mortar. 
A  good  treatment  for  such  stones  is  to  coat  the  back,  sides  and  beds 
with  lime-mortar,  or,  if  this  is  not  efficacious,  with  plaster-of-Paris. 
Natural  ^^^  stones  should  be  laid  on  their  "  natural  beds  "  ; 
bed.     that  is,  in  the  same  position  as  taken  from  the  quarry. 
This  will  bring  the  layers  of  each  stone  into  horizontal  positions,  on 
top  of  each  other,  and  avoid  the  "  peeling  "  so  frequently  seen.    Ash- 
lar should  be  well  anchored  to  the  backing.     The  joints  should  be 
filled  with  putty-mortar,  and  should  be  sufficiently  large  to  take  up 


STONE-WORK. 


123 


the  slirinkase  of  the  backing.  Stones  should  not  be 
*'"  °stones.  so  hirsc  as  to  risk  the  danger  of  their  being  improp- 
erly bedded  and  so  breaking.  Professor  llankine  recommends  for 
soft  stones,  such  as  sand  and  lime  stones,  which  will  crush  with  less 
than  5000  pounds  pressure  per  square  inch,  that  the  length  shall  not 
exceed  three  times  the  depth,  nor  the  breadth  one-and-a-half  times 
the  depth.  For  hard  stones,  which  will  resist  5000  pounds'  compres- 
sion per  square  inch,  he  allows  the  length  to  be  from  four  to  five 
times  the  depth,  and  the  breadth  three  times  the  depth.  Stones  are 
sometimes  joined  with  "rebated"  joints,  or  "  dove-tail"  joints,  the 
latter  particularly  in  circular  work,  such  as  domes  or  light-houses. 

j,j,gt,  '  All  sills  in  either  stone  or  brick  walls  should  be 
hollow,  bedded  at  the  ends  only,  and  the  centre  part  left  hol- 
low until  the  walls  are  thoroughly  set  and  settled;  otherwise,  as  the 
piers  <To  down,  the  part  between  them,  not  being  so  much  weighted, 
will  refuse  to  set  or  settle  equally  with  them,  and  will  force  up  the 
centre  of  siU  and  break  it.  Where  there  are  lintels  across  openings 
in  one  piece,  with  central  muUion,  the  lintel  should  cither  be  jointed 
on  the  muUion,  or  else  the  mullion  bedded  in  putty  at  the  top.  Oth- 
erwise, the  lintel  will  break  ;  or,  if  it  be  very  strong,  the  mullion  will 
split;  for,  as  the  piers  set  or  settle,  the  lintel  tends  to  go  down  with 
them',  and,  meeting  the  mullion,  must  cither  force  it  down,  or  else 
break  it,  or  break  itself. 

Slip-  Walls  of  uneven  height,  even  where  of  the  same 

Joints,      material,  should  be  connected  to  each  other 
by  means  of  a  slip-joint,  so  as  to  provide  for  the  uneven  j 
shrinkage.   Slip-joints  must  be  so  designed  that  while  they    ^^p 
allow  independent  vertical  movement  to  each  part,  neither 
can  separate  from  the  other  in  any  other  direction.     Fig- 
ures 71  to  73  give  a  few  examples. 

Figure  71  shows  the  plan  of  a  gable-wall  connected  with 
a  lower  wall,  by  means  of  a  slip-joint.  Figure  72  shows  the 
corner  of  a  front  stone-wall  connected  similarly  with  side-  ' 
wall  of  brick.  Figure  73  the  corner  of  a  tower  or  chimney  F'e-  ^ '  • 
connected  with  a  lower  wall.  The  joint  must  be  built  plumb  from  top 
to  bottom.  Where  the  higher  waU  sets  over  the  tongue  above  lower 
waU,  one  or  two  inches  must  be  left  hollow  over  the  tongue,  to  allow 
for  settlement  or  shrinkage  of  the  higher  wall  and  to  prevent  its  rest- 
ing on  the  tongue  and  possibly  cracking  it  off.  Where  iron  anchors 
are  used  in  connection  with  slip-joints,  they  should  be  so  arranged  as 
to  allow  free  vertical  movement. 


Low 
Vail. 


124 


SAFK   BUILDING. 


Slone 

ffbnT:, 


Brick  6ide 


Such  joints  and .  anchors  most  be  designed  with  reference  to  each 
special  case.  In  stepped-foundations  (on  shelving-i-ock,  etc.),  or  in 
walls  of  uneven  heights  Avhcrc  slip-joints 
are  impracticable,  the  foundations  or  walls 
should  be  built  up  to  each  successive  level 
and  be  allowed  to  set  thorouglily  before 
building  further.  A  hard  and  quick-setting 
cement  should  be  used,  and  the  joints  made 
as  small  as  possible.   In  no  case  should  one 

wall  of  a  building  be  carried  up  much 

•^'2'  72.  higher  than  the  others,  where  slip-joints  are 

not  to  be  used.  When  building  on  top  of  old  work,  clean  same  off 
thoroughly  or  the  mortar  will  not  take  hold  (clinch).  In  summer,  soak 
Old  and  ^^^  old  work  thoroughly.     Where  new  work  has  to 

new  walls,  be  built  against  old  work,  a  slip-joint  should  be  used, 
if  possible,  or  else  a  straight  joint  should  be  used  with  shp-anchors, 

and  after  the  new  work  has  thoroughly  j 

set,  bond-stones  can  be  cut  in.     In  such.        ^^^r  ii  J       Tovy^T 

cases,  the  foundations  should  be  spread      J 

as  much  as  possible,  to  avoid  serious  set- 
tlements.    In  all  work  involving  old  and 
new  walls  combined,  the  quickest  and  hard- 
est-setting cements  should  be  used.  Some- 
times it  is  advisable  not  to  load  walls  until  ^'S-  73. 
they  have  set,  unless  all  walls  are  loaded  alike,  as  the  uneven  weights 
on  green  walls  are  apt  to  crack  them.   All  walls  should  be  well  braced, 
and  wooden  centres  left  in  till  they  have  set  thoroughly. 
Timber                  Timber  of  any  kind,  in  walls,  should  be  avoided, 

In  walls,  if  possible. 

It  should  only  be  used  for  temporary  support,  as  it  is  liable  to  rot, 

shrink,  burn  out,  or  to  absorb  dampness  and 

swell,  in  either   case  causing  settlements  or 

cracks,  even  if  not  endangering  the  wall.     In 

^no  case  bond  a  wall  with  timbers.    Where  it 

ris  necessary  to  nail  into  a  wall,  wooden  i:)lugs 

are  sometimes  driven  into  the  joints  ;  they  are 

very  bad,  however,  and  liable  to  shake  the  wall 

in  driving.   Wooden  strips,  let  in,  weaken  the 

Fig.  74.  wall  just  that  much.     Wooden  nailing-blocks, 

though  not  much  better,  are  frequently  used.     The  block  should  be 

the  full  thickness  of  the  bricks,  plus  the  upper  and  lower  joints.     If 


TIMnEU    IN    WALLS. 


125 


there  is  any  mortar  over  or  under  tlic  lilock,  the  nailing  will  jar  it 
loose  anil  the  block  fall  out.  The  best  arrangement  to  secure  nail- 
ings  is  to  build-in  porous  ten-a-cotta  blocks  or  bricks. 

Beam  Where  ends  of  beams  are  built  into  the  •wall,  thev 

ends,      should  always  be  cut  off  to  a  slant,  as  shown  in  Fig- 
ure 74. 

The  anchors  should  be  attached  to  the  side,  so  as  to  allow  the  beam 
to  fall  out  in  case  it  is  burned  through.  If  the  beams  were  not  cut  to 
a  slant,  the  leverage  produced  by  their  weight,  when  burned  through, 
would  be  apt  to  throw  the  wall ;  as  it  is,  each  beam  can  fall  out  easily 
and  the  wall,  being  corbelled  over  the  beam-opening,  remains  stand- 
ing. It  is  desirable  to  "  build-in  "  the  ends  of  wooden  beams  as  little 
as  possible,  to  prevent  dry-rot;  if  it  can  bo  arranged  to  circulate  air 
around  their  ends,  it  will  help  preserve  them.  Beams  should  always 
be  levelled-up  with  good-sized  pieces  of  slate,  and  not  v.-ith  wood- 
chips,  which  are  liable  to  crush.  The  old-fashioned  way  of  corbelling 
out  to  receive  beams,  leaving  the  wall  intact,  has  much  tp  commend 
it.  A  modern  practice  is  to  corbel  out  one  course  of  brick,  at  each 
ceiling-level,  just  sufRcicnt  to  take  the  projection  of  furring-strips; 
this  will  stop  draughts  in  case  of  fire,  also  rats  and  mice  from  ascend- 
ing. All  slots  for  pipes,  etc.,  should  be  bricked  up  solid  aro'ind  the 
pipes  for  about  one  foot  at  each  ceiling-level  for  the  same  purposes. 
Wooden  Where  wooden  lintels   are  used  in  walls,   there 

lintels,  should  always  be  a  relieving-arch  over  them,  so  ar- 
ranged that  it  would  stand,  even  if  the  lintel  were  burned  out  o»-  re- 
moved ;  the  lintel  should  have 

'■■■■■'       '        '       ^       ' as  little   bearing  as   possible, 

and  be  shaved  off  at  the  ends. 
Figure  75  shows  a  wooden 
lintel  correctly  built-in.  Figure 
76  shows  a  very  blundering 
way  of  building-in  a  wooden 
lintel,  but  one,  nevertheless, 
^'S«  75.  frequently  met  with.     It  is  ob- 

vious, however,  in  the  latter  case,  that  if  the  lintel  were  removed 
the  abutment  to  the  arch  would  sink  and  let  the  arch  down.  The 
relieving-arch,  after  it  has  set,  should  be  strong  enough  to  carry  the 
wall,  the  lintel  being  then  used  for  nailing  only.  The  rule  for  lin- 
Bonded  tels  is  to  make  their  depth  about  one-tenth  of  the 

arches  best.  span.      Arches  are  built  of  "  row-locks  "  (that   is, 
"  headers,")  or  of  "  stretchers,"  or  a  combination  of  both,  according 


126 


SAFE    BUILDING. 


1      1  '  1      1     1      1            1      1 

1      1      1      [      1      1      1      1      1      1 

1  '  1  '  1     1     (     1     1      1           1 

i       1       I       I      II       1       I'll 

II    tVAvVVVV-v'VSryvyvV' 

7P         1 

(      \^ 

SSSSSSSSIOUM 

y   1      1 

1     '   \ 

/ '     ' 

1       1 

I 

II 

II 

11 

1  '  ^■ 

1      1 

1       1 

1 

1  '   l- 

i      1 

1 

to  design.  The  strongest  arch,  however,  is  one  which  has  a  com- 
bination of  both  headers  and 
stretchers  ;  that  is,  one  which 
is  bonded  on  tlie  face,  and 
also  bonded  into  the  backing. 
Straight  arches  and  arches  built 
in  circular  walls  should  always 
be  bonded  into  the  backing,  or 
if  the  design  does  not  allow  of 
this,  they  should   be   anchored  f^'g-  76. 

back.    "  Straight  "  arches  should  be  built  with  a  slight  "  camber  "  up 

I  towards  the  centre,  to 

allow  for  settlement 
and  to  satisfy  the  eye. 
About  one-eighth  .inch 
rise  at  the  centre  for 
each  foot  of  span  is 
sufficient.  Straight 
pig.  77.  arches  should  never  be 

built,  as  shown  in  Figure  77,  and  known  as  the  French  or  Dutch 
arch,  as  there  is  absolutely  no  strength  to  them.  Fireplaces  are  fre- 
quently arched  over  in  this  way,  but  the  practice  is  a  very  bad  one. 

Brick  facings,  as  laid  up  in  this 
country,  usually  consist  of  all  stretch- 
ers. Every  fifth  or  sixth  course  is 
bonded  into  the  backing,  either  by 
splitting  the  brick  in  two,  as  shown 

^rXcv.  i'l    Figure    78,    and 

facings,  ^^gjj^g  gjjQj.^  headers 

behind   it,  or  by   breaking   off   the 
rear  corners,  as  shown  in  Figure  79,  and  using  diagonally-laid  bond- 
brick. 

The  latter  course  is  the  better,  but  there  is  no  strength  in  either ; 
particularly  as,  as  a  rule,  the  front  brick  are  so  much  softer  and 
weaker  than  those  in  the  backing. 

The  English  bond,  in  which  a  course  of  headers  alternates  with  a 
course  of  stretchers,  is  much  to  be  preferred ;  or,  better  yet,  the 
Flemish  bond,  where  in  each  course  a  header  alternates  with  a 
stretcher.  Of  course,  in  both  EngUsh  and  Flemish  bond,  if  the  front 
brick  are  thinner  than  the  bricks  used  in  the  backing,  larger  and 
more  unsightly  joints  will  be  necessary  in  front. 


Figs.  78  and  79. 


BOND-STONES.  127 

Re-^ular  work  ^^  ^^  ^^^'^>  ^^  ^  ''"^^'  ^^^  ^'^  count  on  the  front  work 

the  best.  fo,.  strength.  We  fretiuently  sec  masons  laying  up 
bi'ick  walls  by  first  hiying  a  single  course  of  Iieaders  or  stretchers  on 
the  outside  of  the  wall,  and  then  one  on  the  inside,  and  then  filling 
the  balance  of  wall  with  bats  and  all  kind  of  rubbish.  This  makes  a 
very  poor  wall.  The  specification  should  provide  that  no  bats  or 
broken  brick  will  be  allowed,  leaving  it  to  the  architect's  discretion 
to  stop  their  use,  if  it  is  being  overdone ;  of  course,  some  few  will 
have  to  be  used.  But,  after  all,  the  best  wall  is  that  one  which  is 
built  the  most  regularly  and  with  the  most  frequent  bonds,  and  no 
architect  should  be  talked  out  of  good,  regular  work,  as  being  too 
theorelical,  by  so-called  "  practical "  men.  The  necessity  for  regular- 
ity  and  bond  is  easily  illustrated  by  taking  a  lot  of  bricks  of  different 
sizes,  or  even  toy  blocks,  and  attempting  to  pile  them  up  without 
regularity ;  or,  even  if  piled  regularly,  without  bond.  It  will  quickly 
be  seen  that  the  most  regular  and  most  frequently-bonded  pile  will 
go  the  highest.  By  "  bond  "  is  meant  alternating  headers  and  stretch- 
ers with  regularity,  and  so  as  to  cover  and  break  joints. 
Use  cf  bond-  "^^^  ^^^^  '^^  "  bond-stones  "  at  intervals  onbj  is  bad ; 

stones,  they  should  be  carried  through  the  whole  surface 
(width  and  length)  of  wall  and  be  of  even  thickness,  or  else  be  omitted. 
Using  bond-stones  in  one  place  only  tends  to  concentrate  the  compres- 
sion on  one  part  of  the  walL     Thus,  bond-stones  built  under  each 
r  other  at  regular  intervals,  as 

,JL    , shown  in  Figure  80,  are  bad, 

V"     '  '\  as  they  give  the  pressure   no 

—  i'  '■  chance  to  spread,  but  keep  con- 

'      I  ^■•  ;■ 'I  ~~-   centrating  it  back  onto  the  part 

|i  'i  of    wall    immediately    under 

.1                   'i,  bond-stones,  whereas,  in    Fi<r- 
|j I.                '  ° 

l'-'v-":.J  ~"~  "^^^^  ^^'  ^"^^  pressure  is  allowed 

i'         — S-  to   spread    gradually    over    a 

I  I  -^  larger  area  of  the  wall.  "Where, 


Pig^  80.  however,  a  heavy  girder,  col- 

umn or  other  weight  comes  on 
a  wall,  it  is  distributed  by  means  of  a  large  block,  generally  granite 
or  stone,  or  sometimes  by  a  large  iron  plate. 

The  block  or  plate  should  have  sufficient  area  not  to  crush  the 
brick-work  directly  under  it.  Where  girder-ends  are  built  into 
walls,  it  is  also  desirable  to  build  a  block  over  the  girder-end  as  well 
as  under  the  same.     The  upper  block  prevents  any  part  of  the  wall 


128  SAFE   BUILDING. 

from  resting  on  the  girder  or  being  affected  by  its  shrinkage,  if  of 

wood ;   if  the  girder  is  of  iron  the  upper  block  will  wedge  in  the 

"'r"  girder-end      more 

firml}^,  and  the  gird- 

~  er  will   be  able  to 

carry   more   load. 

i'  'r  L  See  page    57. 

i'  'i        Anchors     for 

fJ-  'i     girders   and  beams 

[I  'i  ~        are  usually  made  of 

i'  .1,  iron    and    of    such 

i'  —  ■  'i  shape   as  to   allow 

the  girder  or  beam 
'^'S'^'^-  to  fall  out  in  case 

of  fire.     Anchors  made  of  iron  are  not  objectionable  in  inner  walls, 
or  where  not  exposed  to  dampness ;  all  iron  should 
anchors,  be  thoroughly  painted,  however,  with  red  lead  or  me- 
tallic paint.     Before  painting,  all  rust  should  be  scraped  off.     Don't 
believe  the  "  practical  "  man  who  says  the  paint  will  stick  better  if  you 
leave  the  rust  on  the  iron  ;  it  will  stick  better  to  the  rust,  no  doubt,  but 
not  to  the  u'on.    For  outside  work  all  iron  should  be  galvanized ;  but  it 
is  better  to  use  copper  for  anchors,  dowels,  clamps,  etc.    All  copings 
should  be  clamped  together,  the  clamps  being  counter-sunk.   Slanting- 
work  and  tracery  should  be  dowelled  together.    Where  iron  is  let  into 
stones,  and  run  with  lead  or  sulphur,  the  iron  is  apt  to  swell  with 
rust  or  heat  and  burst  the  stone.     Dampness  in  walls  is  one  of  the 
lj,.jp.  worst  dangers,  both  on  account  of  frost  and  decay. 

moulds.  AH  exterior  mouldings  and  sills  should  be  projected 
and  have  "  drip-moulds  "  cut  underneath,  to  cause  the  water  to  drop 
or  drip ;  this  will  prevent  considerable  dampness  and  keep  the  out- 
side surface  of  the  wall  from  becoming  dirty,  as  the  dust  lodging  on  top 
of  mouldings  discolors  the  rain-Avater,  and  the  latter,  instead  of  streak- 
ing down  the  wall,  will  drop  off. 
Hollow  Walls  are  frequently  built  hollow  to  prevent  damp- 

walls,  ness,  but  tliis  raises  many  objections.  Shall  the  inner 
or  outer  wall  be  the  thicker?  If  the  outer  wall,  then  all  beams,  etc., 
have  to  be  that  much  longer,  so  as  to  rest  on  the  stronger  part ;  they 
are  Uable  to  transfer  dampness,  and  then,  too,  the  thicker,  and,  con- 
sequently, greater,  part  of  wall  is  exposed  to  dampness.  If  the  inner 
wall  is  thicker,  the  construction,  so  far  as  beams,  etc.,  are  concerned, 


UXDEIU'IXXIXG.  129 

no  doubt  is  better,  but  the  outer  part  is  apt  to  be  destroyed  by  the 
frost.  Then  at  windows  and  doorways  both  must  be  connected,  and 
dampness  is  apt  to  get  through.  It  is  well  to  ventilate  the  air-space 
between  walls,  at  the  bottom  from  inside  and  at  the  top  from  outside. 
Tlie  bottom  of  spaces  should  be  drained.  Tops  of  arches,  or  lintels 
over  openings,  should  be  cemented  and  asphalted  (in  tlie  air-space), 
to  shed  any  dampness  settling  on  them. 

The  outer  and  inner  walls  should  be  fre-  \ 

quently  anchored  together.   Iron  anchors  ^  -^.^^— -  1 

galvanized,  or  copper  anchors  are  best; 
they  should  have  a  half -twist",  as  shown,  ^'^'  ^^' 

to  prevent  water  running  along  them. 

Care  must  be  taken  to  keep  hollow  walls  free  from  hanging  mortar, 
which  will  communicate  moisture  from  one  wall  to  the  other. 

But  hollow  walls  are  not  nearly  so  good  as  solid  walls  with  porous 
terra-cotta  furrings. 

Where  walls  are  coped  with  stone,  there  should  be  damp-courses 
of  slate  or  asphalt  under  same,  and  the  back  side  should  be  flashed, 
to  prevent  dampness  descending.  If  gutters  are  cut  in  stone  cor- 
nices, they  should  be  lined  with  metal,  preferably  copper,  the  outer 
edge  being  let  into  a  raggle  and  run  in  with  lead. 
Underpinning.  When  a  wall,  already  built,  has  to  have  Its  founda- 
tions carried  down  lower,  it  is  called  "  underpinning  the  wall."  Holes 
are  made  through  the  wall  at  intervals  and  through  these  (at  right 
angles  to  the  wall)  are  placed  the  "  needles,"  that  is,  heavy  timbers, 
which  carry  the  weight  of  the  wall.  Where  the  needle  comes  in  con- 
tact with  the  wall,  small  cross-beams  are  laid  on  its  upper  side,  and 
wedged  and  filled  with  mortar,  to  get  a  larger  and  more  even  bearing 
against  the  wall.  At  the  inner  and  outer  ends  of  the  needles  heavy  up- 
right timbers  are  placed  underneath,  running  down  to  the  new,  lower 
level.  The  foot  or  ground  bearing  of  these  timbers  is  formed  by  heavy 
l)lanks  crossing  each  other,  to  spread  the  weight  over  more  ground; 
.wedges  are  driven  under  the  feet  of  the  uprights,  till  the  ends  of  the 
needle  are  forced  up,  and  the  centre  of  the  needle  shows  a  decided 
downward  curve  of  deflection,  indicating  that  the  weight  of  wall  is  on 
the  needle.  Frequently  jack-screws  are  used  in  place  of  wedges,  to  get 
the  weight  onto  the  uprights.  As  soon  as  the  needles  carry  the  weiglit 
of  wall,  the  intermediate  portions  of  wall  are  torn  out  and  the  excavat- 
ing to  the  lower  level  beghis.  If  the  soil  is  loose,  she.ith-piling  must  be 
resorted  to  on  each  side  of  wall.  Frequently  the  feet  of  the  uprights 
are  "  cribbed,"  that  is,  sheath-piled  all  around,  to  hold  the  ground  under 


130  SAFE    BUIT.DIXG. 

tliem  together  and  keep  it  from  comjiressing.  The  new  wall  is  built  up 
from  the  lower  level  between  and  around  the  needles.  On  top  of  the 
new  wall  two  layers  of  dressed-stone  are  placed  filling  up  between 
the  old  and  new  work.  Between  these  stones  iron  wedges  are  driven 
in  opposite  pairs,  one  from  the  inside  and  one  from  the  outside. 
These  wedges  must  be  evenly  driven  from  both  sides  or  the  wall 
might  tip.  These  wedges  are  driven  until  the  weight  of  the  wall  is  on 
them  and  off  the  needles. 

This  is  readily  seen,  for  the  needles  straighten  out  when  relieved  of 
the  load.  The  jack-screws  are  now  lowered  or  the  wedges  under  the 
uprights  eased  up ;  the  uprights  taken  away,  needles  removed,  and  the 
holes  filled  up.  Underpinning  operations  must  be  slowly  and  care- 
fully performed,  as  they  are  very  risky.  If  there  is  any  danger  of  a 
wall  tipping  during  the  operation,  grooves  are  cut  into  the  wall 
and  "  shores  "  or  braces  placed  against  it.  The  feet  of  the  shores  rest 
on  cross-planks,  same  as  uprights,  and  are  wedged  up  to  get  a  secure 
bearing  of  the  top  of  the  shore  against  the  wall.  "Where  the  outside  of 
a  wall  cannot  be  got  at,  "  spring  needles  "  are  used  from  the  inside. 
That  is,  the  one  end  of  the  needle  acts  as  a  lever  and  supports  the  wall, 
while  the  other,  inner  end,  is  chained  and  anchored  down  to  prevent 
its  tipping  up. 

Strength  The  strength  of  a  wall  depends,  of  course,  largely 

of  bricks.  OQ  the  material  used.  A  good,  hard-burned  brick, 
well  laid  in  cement-mortar,  makes  a  very  strong  wall.  To  tell  a  good 
brick,  first  examine  the  color  ;  if  it  is  very  light,  an  orange-red,  the 
brick  is  apt  to  be  soft.  If  the  brick  is  easily  carved  with  a  knife,  it 
is  soft.  If  it  can  be  crushed  to  powder  easily,  it  is  soft.  If  two 
bricks  are  struck  together  sharply,  and  the  sound  is  dull,  the  bricks 
are  poor ;  if  the  sound  is  clear,  ringing,  metallic,  the  bricks  are  good 
and  hard.  If  a  brick  shows  a  neat  fracture,  it  is  a  good  sign ;  a 
ragged  fracture  is  generally  a  poor  sign.  The  fracture  also  shows 
the  evenness  of  the  burning  and  fineness  of  the  material.  A  brick 
that  chips  and  cannot  be  cut  easily  is  a  good  brick.  The  darker  the 
brick,  the  harder  burned.  This,  of  course,  does  not  hold  good  for 
artificially -colored  bricks.  The  straighter  and  more  regular  the  brick, 
the  softer  it  is  (as  a  rule),  as  hard-burning  is  apt  to  warp  a  brick. 

What  has  been  said  of  the  strength  of  bricks  holds  good  of  terra- 
cotta. The  latter  should  be  designed  to  be  of  same  thickness,  if  pos- 
sible, in  all  parts,  and  any  hollows  caused  thereby  must  be  filled-in 
solid.     It  is  best  to  fill-in  the  hollows  with  bricks  and  mortar  several 


ASriIALT.  131 

davs   in  advance,  and  let  the  filling  set,  so  as  to  be   sure  it  will  not 
swell  up  afterwards  and  burst  the  terra-cotta. 
Strength  "^^  judge  of  the  strength  and  durability  of  stones 

of  stones,  ig  a  more  difhcult  matter.  If  the  stone  be  fractured, 
and  presents,  under  a  magnLTyiug  glass,  a  bright,  clear,  sharp  surface, 
it  is  not  likely  to  crumble  from  decay  ;  if  the  surface  is  dull-appear- 
ing and  looks  earthy,  it  is  likely  to  decay.  Of  course,  samples  can  be 
tested  for  their  crushing  and  tensile  strengths,  etc.  And  we  can  tell 
somewhat  of  the  weathering  qualities  by  observing  similar  stones  in 
old  buildings :  much,  however,  depends  whether  the  stones  come  from 
the  same  part  of  the  quarry.  Another  test  is  to  weigh  different  sam- 
ples, when  dry  ;  immerse  them  in  water  for  a  given  period,  say,  twenty- 
four  hours,  then  weigh  them  again,  and  the  sample  absorbing  the 
least  amount  of  water  (in  proportion  to  its  original  weight)  is,  of 
course,  the  best  stone. 

Another  test  is  to  soak  the  stone  in  water  for  two  or  three  days 
and  put  it  out  to  freeze  ;  if  it  does  not  chip  or  crack,  it  will  probably 
weather  well.  Chemical  tests  arc  made  sometimes,  such  as  using  sul- 
phuric acid,  to  detect  the  presence  of  lime  and  magnesia ;  or,  soak- 
ing the  stones  in  a  concentrated  boiling  solution  of  suljihate  of  soda  ; 
the  stones  are  then  exposed  to  the  air,  when  the  solution  crystalizes 
in  the  pores  and  chips  off  particles  of  the  stone,  acting  similarly  to 
frost.  The  stones  are  weighed  before  and  after  the  tests,  the  one 
showing  the  least  proportional  loss  of  weight  being,  of  course,  the 
better. 

If  stones  are  laid  on  their  natural  beds,  however,  little  need  be 
feared  of  the  result,  if  the  stone  seems  at  all  serviceable.  The  main 
dangers  to  walls  are  from  wet  and  frost.  Very  heavy  and  oft-repeated 
vibrations  may  sometimes  shake  the  mortar-joints,  but  this  need  not 
be  seriously  feared,  in  most  cases ;  machinery  may  often  cause  suffi- 
cient vibrations  to  be  unpleasant,  or  even  to  endanger  wood  or  iron 
work,  but  hardly  well-built  masonry.  Of  course,  the  higher  a  build- 
ing is,  the  greater  will  be  the  amount  of  vibrations  and  their  strength. 
For  this  reason  it  is  advisable  to  place  the  heaviest  machinery  on  the 
,  lowest  (ground)  floor.  The  beds  of  such  machinery  should  be  as  far 
Machinery  ^^  possible  from  any  foundations  of  walls,  columns, 

foundations,  etc.,  and  the  beds  should  be  independent  and  isolated 
from  all  other  masonry.  Malo,  in  Le  Genie  Civil,  recommends  the 
use  of  asphalt  for  machinery-foundations,  as  they  take  up  the  vil)rar 
tions  and  noise,  and  are  as  solid  as  masonry,  if  properly  built.  His 
claim  seems  well  founded,  and  has  been  demonstrated  practically ; 


132  SAFK    UUILDING. 

the  asphalt  foundation  not  only  preventuig  vibrations  but  stopping 
the  sound.  A  wooden  form  is  made,  covered  inside  with  well-greased 
paper ;  into  this  are  placed  slightly-conical  shaped  wooden  bars  and 
boxes,  also  covered  with  well-greased  paper,  which  are  secured  in  the 
places  to  be  occupied  by  the  bolts  and  bolt-heads,  and  arranged  for  easy 
withdrawal.  A  layer  of  melted  asphalt  a  few  inches  thick  is  then  poured 
into  the  mould ;  over  this  are  dumped  heated,  perfectly  clean,  sharp, 
broken  stones  and  pebbles,  rammed  solid,  the  pebbles  filling  all  inter- 
stices ;  then  more  asphalt  is  poured  in,  then  another  layer  of  stones  and 
pebbles,  etc.  It  is  claimed  that  this  foundation  becomes  so  solid  that 
it  will  not  yield  enough  to  disarrange  the  smooth  running  of  any  ma- 
chinery, wliile  its  slightly-elastic  mortar,  besides  avoiding  vibrations 
and  noise,  prolongs  very  much  the  durability  and  usefulness  of  the 
machinery.  Two  dangers  must  be  guarded  against ;  viz.,  the  direct 
contact  of  oil  or  heat  with  the  asphalt.  Stationary  drip-pans  guard 
against  the  former,  while  a  layer  of  rubber,  wood,  cement,  or  other 
non-conductive  material  would  accomplish  the  latter  object.  Where 
noise  from  machinerj-  is  to  be  avoided,  a  layer  about  one  inch  thick, 
of  hard  rubber  or  soft  wood,  should  be  placed  immediately  under  the 
en<Tine-plate.  If  this  layer  were  bedded  in  asphalt  the  precaution 
•would  be  still  more  effective. 

^     ,.^      -  In  all  cases  where  asphalt  is  used,  that  with  the 

Quality  of  '■ 

asphalts,  least  proportion  of  bitumen  should  be  preferred. 

Seyssell  asphalt,  which  comes  from  France,  is  undoubtedly  the 
best,  and  next  to  this  comes  the  Swiss  or  Xeuchatel  asphalt. 

Trinidad  asphalt,  which  is  much  used  in  this  country,  is  much  in- 
ferior, being  softer  and  containing  a  larger  proportion  of  bitumen  or 
tar  —  a  great  disadvantage  in  manj'  cases. 

r^  ■  „o  «„«.,  In  all  walls  trv  to  get  all  openings  immediately 
Openmss  over  .  o  i  o 

each  other,  over  each  other.    A  rule  of  every  architect  should  be 
to  make  an  elevation  of  ever)/  interior  wall,  as  well  as  of  the  exterior 
walls,  to  see  that  openings  come  over  each  other. 
Tower  ^^  is  foolish  to  make  chimneys  or  tower-walls  un- 

walls.  necessarily  thick  (and  heavy),  as  they  brace  and  tie 
themselves  together  at  each  corner,  and,  consequently,  are  much 
stronger  than  ordinary  walls.  Tower-walls,  however,  often  require 
thickening  all  the  way  down,  to  allow  for  deep  splays  and  jambs  at 
the  belfry  openings.  The  chief  danger  in  towers  is  at  the  piers  on 
main  floor,  which  are  frequently  whittled  down  to  dangerous  propor- 
tions, to  make  large  door  openings.  In  tall  towers  and  chimneys  the 
leverage  from  wind  must  be  carefully  considered. 


STKKXGTII   01"   WALLS.  103 

Considering  the  importance  of  ascertaining  tlic  exact  strengtli  of 
walls,  it  is  remarkable  that  so  little  attention  has  been  paid  to  the 
subject  by  writers  and  experimenters.     The  only  known  rule  to  the 
writer  is  Rondelet's  graphical  rule,  which  is  as  follows : 
Rondelet's  If  A  B  (Figure  83)  be  the  height  of  a  wall,  B  C 

rule,  the  length  of  wall  without  buttress  or  cross-wall  (AB  C 
being  a  right  angle),  then  draw  A  C  ;  make  A  D  =  to  either  ^  or  ^^ 
or  ^  of  A  B,  according  to  the  /^ 
nature  of  wall  and  building  it  is 
intended  for  {^^  for  dwellings,  J^ 
for  churches  and  fireproof  build- 
ings, and  I  for  warehouses)  ;  then 
make  A  E  =  A  D.  and  draw  E  F 
parallel  to  A  B ;  then  is  B  F  the 
required  thickness  of  wall.  The 
rule,  however,  in  many  cases,  gives 
an  absurd  result.  Gwilt's  "■En- 
cyclopfEdla  of  ArchUecture"  gives  this  rule,  and  many  additional 
rules,  for  its  modification.  There  are  so  many  of  them,  and  they  arc 
so  complex,  however,  as  to  be  utterly  useless  in  practice.  Most 
cities  have  the  thickness  of  walls  regulated  by  law,  but,  as  a  rule, 
these  thicknesses  give  the  minimum  strength  that  will  do,  and  where 
they  do  not  regulate  the  amount  and  size  of  flues  and  openings,  fre- 
quently allow  dangerously-weak  spots  in  the  wall. 
Formula  The  writer  prefers  to  use  a  formula,  whicli  he  has 

for  masonry,  constructed  and  based  on  Rankine's  formula  for  long 
pillars,  see  Formula  (3),  and  which  allows  for  evei'y  condition  of  height, 
load,  and  shape  and  quality  of  masonry.  In  the  case  of  piers,  columns, 
towers  or  chimneys,  whether  square,  round,  rectangular,  solid  or  hol- 
low, the  Formula  (3)  can  be  used,  just  as  there  given,  inserting  for  p^ 
its  value,  as  given  in  the  last  column  of  Table  I,  using,  of  course,  the 
numbered  section  corresponding  to  the  cross-section  of  the  pier,  col- 
umn or  tower.  By  taking  cross-sections  at  different  points  of  the 
height,  and  using  for  I  the  height  in  inches  from  each  such  cross- 
section  to  the  top,  we  will  readily  find  how  much  to  offset  the  wall. 
Care  must  be  taken,  where  there  are  openings,  to  be  sure  to  get  the 
piers  heavy  enough  to  carry  the  additional  load ;  the  extra  allowance 
for  piers  should  be  gotten  by  calculating"  the  pier  first  as  an  isolated 
pier  of  the  height  of  opening,  and  then  by  taking  one  of  our  cross- 
sections  of  the  whole  tower  or  chimney  at  the  level  where  the  open- 
ings are,  and  using  whichever  result  required  the  greater  strength. 


134  SAFE    BUII-DIXG. 

As  it  would  be  awkward  to  use  the  heiglit  I  in 
piers, chimneysinches,  we  can  modify  the  formula  to  use  the  height 
and  towers.     ^  in  feet.    Further,  we  know  the  value  of  n  for  brick- 
work, from  Table  II,  and  can  insert  this,  too ;  we  should  then  havt; : 
For  hrick  or  rubble  piers,  cldmneys  and  toioers,  of  whatever  shape : 


<7) 


*^^Z^  (50) 


T^= ■'  'l-2x  (60) 


l+0,475.(^) 

"Where  to  =  the  safe  total  load  on  pier,  chimney  or  tower  in  pounds. 
"Where  a  =:  the  area  of  cross-section  of  pier,  etc.,  in  square  inches, 
at  any  point  of  height. 

"Where  i  =  the  height,  in  feet,  from  said  point  to  top  of  masonry. 

Where  (  —  \^=  the  safe  resistance  to  crushing,  per  square  inch, 

as  given  in  Table  Y.     (See  page  135.) 

"Where  q^  =  the  square  of  the  radius  of  gyration,  of  the  cross-sec- 
tion, in  inches,  as  given  in  Table  I. 

If  it  is  preferred  to  use  feet  and  tons  (2000  lbs.  each)  we  should 
have 

U  +  0,04G.(|l^) 

"WTiere  "W  =  the  safe  total  load  on  masonry,  in  tons,  of  2000  lbs 
each. 

"Wliere  A  =  the  area  of  cross-section  of  masonry,  at  any  point  of 
height,  in  square  feet. 

"Where  L  =  the  height,  in  feet,  from  said  point  to  top. 

"Where  (--,j=  the  safe  resistance  to  crushing,  in  lbs.,  per  square 

inch,  as  given  in  Table  Y.     (See  page  135.) 

Where  P^  =  the  square  of  the  radius  of  gyration,  as  given  in  last 
column  of  Table  I,  —  but  all  dimensions  to  be  taken  in  feet. 

To  obtain  the  load  on  masonry,  include  weight  of  all  masonry, 
floors,  roofs,  etc.,  above  the  point  and  (if  wind  is  not  figured  sepa- 
rately) add  for  wind  15  lbs.  for  each  square  foot  of  outside  superfi- 
cial area  of  ad  walls  above  the  point. 

Where  towers,  chimneys,  or  walls,  etc.,  are  isolated,  that  is  not 
braced,  and  liable  to  be  blown  over  by  wind,  the  wind-pressure,  must 
be  looked  into  separately. 


BRICKWORK. 


135 


In  regard  to  the  use  of  (  —  ]  the  safe  resistance,  per  square  inch, 


/ 

of  the  material  to  crushing,  it  should  be  taken  from  Table  V. 
that  for  rubble-work  -we  should  use  : 


So 


(7)= 


100 


And  the  same  for  poor  quality  brick,  laid  in  lime  mortar. 

For  fair  brick  in'  lime  and  cement  (mixed)  mortar,  we  should  use : 

(7)=- 

And  for  the  best  brickwork  in  cement  mortar,  we  should  use  : 

If,  however,  a  wall  (or  pier)  is  over  3  feet  thick,  and  laid  in  good 
cement  mortar,  with  the  best  hard-burned  brick,  and  there  are  not 
many  flues,  etc.,  in  the  wall,  we  can  safely  use : 

If  the  wall  (or  pier)  is  over  3  feet  thick,  and  there  are  no  flues  or 
openings,  and  the  best  brick  and  foreign  Portland  cements  are  used, 
it  would  be  perfectly  safe  to  use  : 

Example. 
A  totoer  IG  feet  square  outside,  carries  a  steeple  weigh- '^ 
ing,  including  toind-pressure,  some  15  tons.     The  belfry '00 
openings,  on  each  side,  are  central  and  virtualbj  equal  to    | 
openings  ?>  feet  wide  by  18  feet  high  each. 


There  are  S  +-**^ 


feet  of  solid  loall  over  openings.    \ 

What  should  be  the  thickness    I 

I 


S^  of  belfry  piers  ?     The  mason-    I 

^  ry  is  ordinary  rubhle-work.       'x 

In  the  first  co 

place  we  will   1 


Tower 

Walls. 


try     Formula     (60)     giving   ' 
strength  of  whole  tower  at   ■ 
f%  base  of    belfry   piers.      The  _i  _ 
load    will    be:      4.(2G.16    — 
18.8  — 2(i. If)  =892    superfi-       ^^^ 
cial  feet  of  masonry  20"  thick      Fig.  35. 
and  weighing  250  lbs.  i)cr  superficial  foot  =  223000  lbs.  (see  Figures 


Fig.  84 


136  SAFE   BUILDINa. 

84  and  85)  :  add  to  tliis  spire  and  we  have  at  foot  of  belfry  piers : 
Actual  load  =  ^oSOOO  lbs.  or  =  126  tons. 
XoTv  P2  (the  square  of  the  radius  of  gyration)  would  be  P^  =  --j- ; 

the  area  A:=162  —  (12§2-j-4.8.1f)  =43  ;  the  moment  of  inertia  1 
_  j^.   (1G4_12|*— 3^.83  — 8.1G3+8.12§3)  =1799. 

Therefore  P2=?^  =  41,8.     Xo.v  for    rubble -work,  Table  Y, 

/  ^  '\  =r  100  ;    and,  from   Formula  (GO),  the  safe  load  would  be  : 
^  43.100  4300 


18,"8;  or,  say,  ?/  =  19". 


,  -r  ^'^=14  +  0,046.2^        1^''' 

*^         -^  ^  41,8 

-^  l^  ^  291  tons;  or  more  than  strong  enough. 

M :^ rz'^  '■'       Piers  at  Now  let  us  examine  the 

^  >^  opening,  strength  of    each  pier  by 

~\)'APs-   -^  itself.  Figure  86.     In   the    first  place  we 

Pj     gg^  must  find  the  distance  y  of  the  neutral  axis 

M-N  from  say  the  line  A  B.    This  from 
Table  I,  Section  Xo.  20  is  : 

ig|l'+.20..S.(20  +  ¥)_ 
y  ~  48.20  +  20.28 

Now  i  =  ?M^i+^5^i^±^^^'=   272347      (in     inches)      and 

a  =  20.48  -|-  20.28  =:  1520  square  inches,  therefore  q^  =  —  =  179 

(in  inches). 

The  len^i-th  of  each  pier  is  IS  feet,  or 

'  L=18. 
Therefore,  from  Formula  (59)  we  have  the  safe  load : 
^^_       1520.100 _  gjg^Q  pounjs^ 

,       ,      r^    ,  ~-    18.18 

1  +  0,4-0.^^ 
or  say  the  safe  load  on  each  pier  would  be  41  tons. 

we  know  is  — ^=:  31 1  tons,  or 
4 

Xow  let  us  see  how  far  down  it  would  be  safe  to 
of  walls,  carry  the  20"  walls.     We  use  formula  (60)  and  have 
from  Section  Number  4,  of  Table  1  : 

P2  ^16^+l.!r  =,  34|  (in  feet). 


126 
:ual  load  we  Jinow  is 

than  safe 


The  actual  load  we  know  isi+-=  31  ^  tons,  or  the  pier  is  more 

4 


TOWER  WALLS.  137 

The  area  would  be 

A  =  1G2— 12|2  =  9G  square  feet. 
The  load  for  each  additional  foot  under  belfry  would  be  then  : 

90.150  =  14400  lbs.,  or  7,2  tons. 
The  whole  load  from  top  down  for  each  additional  foot  would  be, 
in  tons : 

W.  =  12G  +  (L  —  26).7,2  =  7,2.L  —  61 
While  the  safe  load  from  Formula  (60)  would  be : 

IV  = ^li^O 

14  +  0,046.3^ 

Now  trying  this  for  a  point  50  feet  below  spire,  we  should  have 
the  actual  load : 

W.  =  7,2.50  —  61  =  299  tons. 
and  the  safe  load : 

W=  ^G-'i^^O 554  toTjg    or,   we  can  go  still 

14+0,046.^ 

lower  with  the  20"  work.     For  70  feet  below  spire,  we  should  have 
actual  load : 

W,=  7,2.70  —  61  =443  to 
while  the  safe  load  : 

W—  ^"^--^^^ =468  tons,  or,  70  feet  would  be 

70.70 
14X0,046.1|^ 

about  the  limit  of  the  20"  work. 

If  we  now  thicken  the  walls  to  24",  we  should  have 

A=  112  square  feet. 

P2  from  Section  4,  Table  I,  =  33^  (in  feet). 
The  weight  per  foot  would  be  112.150  =  16800  lbs.  (or  8,4  tons) 
additional  for  every  foot  in  height  of  24"  woi-k. 
Therefore  the  actual  load  would  be, 

443  -f  (L—  70).8,4  or 

"W.  =  Z.8,4— 145. 
Now,  for  X=:  80  feet,  we  should  have  the  actual  load : 

W,  =  527  tons,  while  the  safe  load  would  be : 

W=—^^^^ =  491  tons. 

U  +  0,046.«ig? 

This,  though  a  little  less  than  the  actual  load,  might  be  passed. 
Rubble  stone  work,  however,  should  not  be  built  to  such  height,  good 


138  SAFE   BUILDING. 

brickwork  in  cement  would  be  better,  as  it  can  be  built  ligbter ;  for 
/'£_\_200,  would  give  larger  results,  and  brickwork  weigbs  less, 
besides  ;  then,  too,  we  have  the  additional  advantage  of  saving  con- 
siderable weight  on  the  foundations. 

Thickening  the  walls  of  a  tower  or  chimney  on  the  inside  does 
not  strengthen  them  nearly  so  much  as  the  same  material  applied  to 
the  outside  would,  either  by  offsetting  the  wall  outside,  or  by  build- 
ing piers  and  buttresses. 

It  is  mainly  for  this  reason,  and  also  to  keep  the  flue  uniform,  that 
chimneys    have    their  outside  dimensions  increased    towards    the 

bottom. 

Example. 

Calculation  of        A  circular  brick  chimney  is  to  he  built  150  feet  high, 
chimneys,  ^/^g  y^„g  entering  about  6  feet  from  the  base;  the  horse- 
poioer  of  boilers  is  1980  HP.    What  size  should  the  chimney  be? 
The  formula  for  size  of  flue  is : 

_^^0,3.//P+10  (61) 

VL 
Where  A  =  the  area  of  flue,  in  square  feet. 
Where  L  =  the  length  of  vertical  flue  in  feet. 
Where  HP =the  total  horse-power  of  boilers. 
Size  of  flue.       A  circular  flue  will  always  give  a  better  draught 
than  any  other  form,  and  the  nearer  the  flue  is  to  the  circle  the  bet- 
ter will  its  shape  be. 

In  our  case  the  flue  is  circular,  so  that  we  will  have 

A=^.E^  (see  Table  I,  Sec.  No.  7)  or 

V    22 
Inserting  the  value  of  A  from  formula  (61)  we  have: 
R=     /7     0,3.1080 -f  10^      M.  50,3  =  4 
V22-        VUT  V^^ 

or  the  radius  of  flue  will  be  4  feet  (diameter  8  feet). 

Now  making  the  walls  at  top  of  chimney  8"  thick  and  adopting 
the  rule  of  an  outside  batter  of  about  f  to  the  foot,  or  say  4"  every  15 
feet,  we  get  a  section  as  shown  in  Figure  87. 

Let  us  examine  the  strength  of  the  chimney  at  the  five  levels 
A,  B,  C,  D  and  E. 

The  thickness  of  the  base  of  each  part  is  marked  on  the  right-hand 
section,  and  the  average  thickness  of  the  section  of  the  part  on  the 
left-hand  side. 


CHIMNEY  WALLS. 


189 


Take  the  part  above  A ; 
the  average  area  is  (^^.5^ 
— flue  area)  or,  78  —  50  = 
28  square  feet. 

This  multiplied  by  the 
height  of  the  part  and  the 
■weight  of  one  cubic  foot 
of  brickwork  (112  lbs.) 
"■ives  the  weight  of  the 
■whole,  or  actual  load. 

W.  =  28.30.112  = 
94080  lbs.,  or  47  tons. 

The  area  of  the  base  at 
A  would  be : 

A  =(=^.51^  —  flue 

area);  or  A  =  89  —  50  = 

o9  square  feet. 

The  height  of  the  part 
is  2:  =  30. 

The  square  of  the  ra- 
dius of  gyration,  in  feet, 

''p2=5iMll^_  11,11 

4 

Inserting  these  values 
in  Formula  (GO)  the  safe 
load  at  A  would  be : 
39.200 

=  440  tons,  or  about  nine 
times  the  actual  load. 
Now,  in  examining  the 
joint  B  we  must  remem- 
ber to  take  the  whole  load 
of  brickwork  to  the  top 
as  well  as  whole  length 
L  to  top  (or  CO  feel). 
The  load  on  B  we  find  is : 
W,  =  131  tons,  while 
the  safe  load  is : 


vO/-zMf  itfTK/iSCZ 


Fig.    B7. 


140  SAFE   BUILDING. 

n.=  _5t?«»  =472t„n.. 

14  +  0,046.5H2 

Similarly,  we  should  find  on  C  the  load : 

W,  =  259  tons,  while  the  safe  load  ia : 
jy.  89.200  ...  .  „„ 

^    '        15,11 
On  D  we  should  find  the  load  : 

"W,  =  432  tons,  while  the  safe  load  is  : 

W= '-1^:^ =  448  ton«. 

14  +  0,046.^^:1^ 
~    '  17,44 

Below  D  the  wall  is  considerably  over  three  feet  thick,  and  is  soUd, 

therefore  we  can  use  (  — r  )  =  300,  provided  good  Portland  cement 

is  used  and  best  brick,  which  should,  of  course,  be  tlie  case  at  the 
base  of  such  a  high  chimney.     We  should  have  then  the  load  on  E  : 

W,=  657  tons,  while  the  safe  load  is  : 

W= 151:300 ^  ggQ  ^ 

14  +  0,046.1^^ 

The  chimney  is,  therefore,  more  than  amply  safe  at  all  points,  the 
bottom  being  left  too  strong  to  provide  for  the  entrance  of  ilue, 
which  will,  of  course,  weaken  it  considerably.  "We  might  thin  the 
upper  parts,  but  the  bricks  saved  would  not  amount  to  very  much 
and  the  offsets  would  make  very  ugly  spots,  and  be  bad  places  for 
water  to  lodge.  If  the  chimney  had  been  square  it  would  have  been 
much  stronger,  though  it  would  have  taken  considerably  more  mate- 
rial to  build  it. 

It  is  generally  best  to  build  the  flue  of  a  chimney  plumb  from  top 

to  bottom,  and,  of  course,  of  same  area  throughout.     Sometimes  the 

flue  is  gradually  enlarged  towards  the  top  for  some  five  to  ten  feet 

in  height,  which  is  not  objectionable,  and  the  writer  has  obtained 

good  results   thereby ;   some  writers,  though,  claim  the  flue  should 

•w  r  #»i.i        be  diminished  at  the  top,  which,  however,  the  writer 

Tops  Of  Chlm-  '■  \ 

ney  Flues.  has  never  cared  to  try.  Galvanized  iron  bands 
should  be  placed  around  the  chimney  at  intervals,  particularly  around 
the  top  part,  which  is  exposed  very  much  to  the  disintegrating  effects 
of  the  weather  and  the  acids  contained  in  the  smoke.  Xo  smoko  flue 
should  ever  be  pargetted  (plastered)  inside,  as  the  acids  in  the  smoke 
will  eat  up  the  lime,  crack  the  plaster,  and  cause  it  to  faU.     The 


BULGING   OF   WALLS. 


141 


crevices  will  fill  with  soot  and  Ije  liable  to  catch  fire.  The  mortar- 
joints  of  flues  should  bo  of  cement,  or,  better  yet,  of  fire-clay,  and 
should  be  carefully  struck,  to  avoid  being  eaten  out  by  the  acids. 

.      Where  walls  arc  long,  without  buttresses  or  cross- 
Calculation    of  ,         ,,        . ,  n        r    i     m  r    « 

Walls.- Bulging,  walls,  such   as  gable-walls,  side-walls   ot    building^ 

etc.,  we  can  take  a  slice  of  the  wall,  one  running  foot  in  length,  and 

consider  it  as  forced  to  yield  (bulge)  inwardly  or  outwardly,  so  that 

for  p^  we  should  use : 

Q-  _  i^ ;  where  d  the  thickness  of  wall  in  inches.     The 
^  12 

area  or  a  would  then  be,  in  square  inches,  a  =  \1.d. 
Inserting  these  values  in  formula  (59)  we  have  for 


BRICK   OR   STONE   WALLS. 

fl (62) 


(t) 


M7  =  - L3 

0,0833  +  0,475.-^ 
Where  w  =  the  safe  load,  in  lbs.,  on  each  running  foot  of  wall 

(d"  thick). 

Where  J  =  the  thickness,  in  inches,  of  the  wall  at  any  point  of  its 

lieight. 

Where  L=the  height,  in  feet,  from  said  point  to  top  of  wall. 

Where  (-^  )  =  the  safe  resistance  to  crushing,  in  lbs.,  per  square 

inch,  as  given  in  Table  V.     (See  page  135.) 

If  it  is  preferred  to  use  tons  and  feet,  we  insert  in  formula  (GO)  : 
for  A  =  D,  where  D  the  thickness  of  wall,  in  feet,  and  we  have : 

P2=±--:  therefore 
12  ' 

o 

U  +  0,552.jp 
Where  AV  =  the  safe  load,  in  tons,  of  2000  lbs.,  on  each  running 
foot  of  wall  (D  feet  thick). 
Where  D  =  the  thickness  of  wall,  in  feet,  at  any  point  of  its  height. 
Where  L  =  the  height,  in  feet,  from  said  point  to  the  top  of  wall. 
"Where  (-^)  =  the  safe  resistance  to  crushing  of  the  material, 

in  lbs.,  per  square  inch,  as  found  in  Table  V.     (See  page  135.) 
Anchored  Walls.     AVliere  a  wall  is  thoroughly  anchored  to  each  tier 


142 


SAFE   BUILDING. 


of  floor  beams,  so  that  it  cannot  possibly  bulge,  except  between  floor- 
beams,  use  the  height  of  story  (that  is,  height  between  anchored 


+-- 


'  QrfY  rLOon. 


ID 


+-■ 


;  7 


^ 


2& 


1 


i    5 


■!o 


32 


36 


-In 


40 


r 


F/krmooJi 


Fig.  88.  Fig.  89.  Fig.  90. 

beams  in  feet)  in  place  of  L  and  calculate  d  or  D  for  the  bottom  of 
•wall  at  each  story. 

The  load  on  a  waU  consists  of  the  wall  itself,  from  the  point  at 
which  the  thickness  is  being  calculated  to  the  top,  plus  the  weight  of 


EXAMPLES  OK   WALLS.  143 

one  foot  in  widtli  by  half  the  span  of  all  the  floors,  roofs,  partitions, 
etc.  Where  there  are  openings  in  a  wall,  add  to  pier  the  proportion- 
ate weight  Avhich  would  come  over  opening ;  that  is,  if  we  find  the 
load  per  running  foot  on  a  wall  to  be  20000  lbs.,  and  the  wall  con- 
sists of  four-foot  piers  and  three-foot  openings  alternating,  the  piers 
will,  of  course,  carry  not  only  20000  lbs.  per  running  foot,  but  the 
60000  lbs.  coming  over  each  ojoening  additional,  and  as  there  are 
four  feet  of  pier  we  must  add  to  each  foot  ^^^i^z=  15000  lbs. ;  ^vc 
therefore  calculate  the  pier  part  of  Avail  to  carry  35000  lbs.  per  run- 
ning foot.  The  actual  load  on  the  wall  must  not  exceed  the  safe 
load  as  found  by  the  formula  (G2)  or  (03). 

Example. 
Wall  of  Country     A  two-Hlorij-and-atlic   dioelling  has  hriclc  icalls  12 
^"®®"         inches  thick;  the  walls  carry  two  tiers  of  heanis  of 
20  feet  span;   is  the  wall  strong  enough  f     The  brickwork  is  good  and 
laid  in  cement  mortar. 

We  will  calculate  the  thickness  required  at  first  story  beam  level, 
Figure  88. 

The  load  is,  pei  running  foot  of  wall : 

Wall  =22.112  =  2464  lbs. 

Wind  =22.  15=    330  lbs. 

Second  floor  =10.90=    900  lbs. 

Attic  floor  =  10.  70  =    700  lbs. 

Slate  roof  (inch  wind  and  snow)  =  10.  50  =  500  lbs. 
Total  load  =4«94  lbs. 

For  the  quality  of  brick  described  we  should  take  from  Table  V  : 


('-^)  =  200  lbs. 


J 

The  height  between  floors  is  10  feet,  or 
L=10, 
therefore,  using  formula  (02)  we  have  : 

d.\ 


Kj) 


12-200  =35807  lbs. 


0,0833  +  0,475.  ^^      0,0833  +  0,475. 1M2 

So  that  the  wall  is  amply  strong.  If  the  wall  were  pierced  to  the 
extent  of  one-quarter  with  openings,  the  weight  per  running  foot 
would  be  increased  to  6525  lbs.  Over  700  lbs.  more  than  the  safe 
load,  still  the  wall,  even  then,  would  be  safe  enough,  as  we  have 
allowed  some  330  lbs.  for  wind,  which  would  rarely,  if  ever,  be  so 
strong;  and  further,  some  1200  lbs.  for  loads  on  floors,  also  a  very 


144  SAFE    UUILDING. 

ample  allowance;  and  even  if  the  two  ever  did  exist  together  it 

would  only  run  the  compression  T—  j  up  to  225  lbs.  per  inch,  and 

for  a  temporanj  stress  this  can  be  safely  allowed. 

The  writer  would  state  here,  that  the  only  fault  he  finds  with  for- 
mulae (59),  (60),  (62)  and  (63),  is  that  their  results  are  apt  to  give 
an  excess  of  strength ;  still  it  is  better  to  be  in  fault  on  the  safe  side 
and  be  sure. 

Example. 

Walls  of  City  ■^'''^  hrick  walls  of  a  warehouse  are  115  feel  high, 

Warehouse,  the  8  stories  are  each  14  feet  high  from  floor  to  floor, 
or  12  feet  in  the  clear.  The  load  on  floors  per  square  foot,  including 
the  fire-proof  construction,  loill  average  300  lbs.  What  size  should  the 
walls  be  f  The  span  of  beams  is  26  feet  on  an  average.  (See  Fig.  89, 
page  142.) 

According  to  the  New  York  Building  Law,  the  required  thicknesses 
would  be:  first  story,  32" ;  second,  third,  and  fourth  stories,  28"; 
fifth  and  sixth  stories,  24" ;  seventh  and  eighth  stories,  20". 

At  the  seventh  story  level  we  have  a  load,  as  follows,  for  each  run- 
ning foot  of  wall : 

Wall  =30.1|.112  =  5600 

Wind  =        3.0.30=    900 

Eoof  =      13.120=7  1560 

Eighth  floor   =      13.300  =  3900 

Total  =  11960  lbs.,  or  6  tons. 

The  safe  load  on  a  20"  wall  12  feet  high,  from  formula  (63)  is: 
W—  l|-.200  333 

-  .^,   ,33,  12.12  =  14  +  0,552.51,84  =  ^'^^^  *°"^'  °^ 

'       'If-lf 
15622  lbs. 

If  one-quarter  of  the  wall  were  used  up  for  openings,  slots,  flues, 

etc.,  the  load  on  the  balance  would  be  8  tons  per  running  foot,  which 

IS  still  safe,  according  to  our  formula. 

At  the  fifth-story  level  the  load  would  be : 

Load  above  seventh  floor  =  11960 

Wall  =28.2.112=1:    6272 

Wind  =28.30      =      840 

Sixth  and  seventh  floors  =  2.13.300  =    7800 

Total  =  26872  lbs.  or  13i 

tons. 


WAREHOUSE   WALL.  145 

The  safe  load  on  a  24"  wall,  12  feet  higli,  from  formula  (G3)  is : 
w  2.200  400 

14  +  0,5o2 '      ' 

^    '  2.2 

or  23C18  lbs. 

This  is  about  10  per  cent  less  than  the  load,  and  can  be  passed  as 
safe,  hut  if  there  -were  many  ilucs,  openings,  etc.,  in  wall,  it  should  be 
thickened. 

At  the  second-story  level  the  load  would  be  : 

Load  above  fifth  floor  =  26872 

Wall  =42.2^.112  =  10976 

Wind  =  42.  30         =    1260 

Third,  fourth  and  fifth  floors  =    3.13.300    =  11700 

Total  =  50808,    or 

25  tons. 

The  safe-load  on  a  28"  wall,  12  feet  high,  from  formula  (63)  is : 

,Tr  2i.200  467  ,c  oo  * 

VV  = ^ ; =  16,33  tons, 

14 -I- 0  559  IM?  ""  14  +  0'552.26,45 

or  32660  lbs. 

Or,  the  wall  would  be  dangerously  weak  at  the  second-floor  level. 
At  the  first-floor  level  the  load  would  be  : 

Load  above  second  floor      =  50808 
Wall  =  14.  2|.112  =    4181 

Wind  =  14.30  =      420 

Second  floor  =  13.300         =     3900 

Total  =  59309,    or  29^  tons. 

The  safe-load  on  a  32"  wall,  12  feet  high,  from  formula  (63)  is: 

W  = '-^^ r-^^^ 21,169  tons, 

14-f0,552.1^2~l^  +  «'-5^-20,25- 

'  2f.2|-  or  42338  lbs. 

The  wall  would,  therefore,  be  weak  at  this  point,  too. 

Now  while  the  conditions  we  have  assumed,  an  eight-story  ware- 
house with  all  floors  heavily  loaded,  would  be  very  unusual,  it  answers 
to  show  how  impossible  it  is  to  cover  every  case  by  a  law,  not  based 
on  the  conditions  of  load,  etc.  In  reality  the  arrangements  of  walls, 
as  required  by  the  law,  are  foolish.  Unnecessary  weight  is  piled  on 
top  of  the  wall  by  making  the  top  20"  thick,  which  wall  has  nothing 
to  do  but  to  carry  the  roof.  (If  the  span  of  beams  were  increased 
to  31  feet  or  more  the  law  compels  this  top  wall  to  be  24"  thick,  if  41 
feet,  it  would  have  to  be  28"  thick,  an  evident  waste  of  material.)  It 


146 


SAFK    BUILDING. 


■would  be  inucli  better  to  make  the  top  walls  lighter,  and  add  to  the 
bottom ;  in  tliis  case,  the  writer  would  suggest  that  the  eighth  story 
be  12"  ;  the  seventh  story  IG"  ;  the  sixth  story,  20";  the  fifth  story, 
24";  the  fourth  story,  28";  the  third  story,  32";  the  second  story, 
3G",  and  the  first  story  40",  see  Figure  90.     (Page  142.) 

This  would  represent  Jmt  4-|  cubic  feet  of  additional  brickwork  for 
every  running  foot  of  wall ;  or,  if  we  make  the  first-story  wall  3(i" 
too,  as  hereafter  suggested,  the  amount  of  material  would  be  exactly 
the  same  as  required  by  the  law,  and  yet  the  wall  would  be  much 
better  proportioned  and  stronger  as  a  whole.  For  we  should  find 
(for  L  =  12  feet), 

Actual  load  at  eighth-floor  level,  3832  ) 

Safe  load  on  a  12"  wall  from  Formula  (G2)         42D8  \ 

Actual  load  at  seventh-floor  level,  10243  ) 

Safe  load  on  a  16"  wall  from  Formula  (G2)         9135  | 

Actual  load  at  sixth-floor  level,  1717G  ) 

Safe  load  on  a  20"  wall  from  Formula  (G2)        15729  ^ 

Actual  load  at  fifth-floor  level,  24G32  [ 

Safe  load  on  a  24"  wall  from  Formula  (G2)        23750  ) 

Actual  load  at  fourth-floor  level,  32G11  } 

Safe  load  on  a  28"  wall  from  Formula  (G2)        32825  j 

Actual  load  at  third-floor  level,  41112 

Safe  load  on  a  32"  wall  from  Formula  (62)        42638 

Actual  load  at  second-floor  level,  5013G 

Safe  load  on  a  3G"  wall  from  Formula  (62)       5290 

Actual  load  at  first-floor  level,  69683 

Safe  load  on  a  40"  wall  from  Formula  (62)       6349 

The  first-story   wall  could 

safely  be   made   36"   if  the 

brickwork  is  good,  and  there 

are  not  many  flues,  etc.,  in 

walls,  for  then  we  could  use 


(7) 


.  =  250,    which    would 

give  a  safe  load  on  a  36" 
wall  =  65127  lbs.,  or  more 
than  enough. 

The  above  table  shows  how 
very  closely  the  Formula  (62) 
would  agree  with  a  joractical 
and  common  -  sense  arrange- 
ment of  exactly  the  same 
amount  of  material,  as  required  by  the  law. 


56) 
)2| 

)2; 


Fig.  91. 


TIIUUST    OK    DAKUKLS. 


147 


_.        ^    ,  Now,  if  the  upper  (loor  were  liulen  with  l);uTel«, 

Thrust  of  »  ' 

barrels,  there  might  be  some  danger  of  these  thrusting  out  the 
wall.  We  will  suppose  an  extreme  ease,  four  layers  of  flour  barrels 
l)aekeil  against  the  wall,  leaving  a  5-foot  aisle  in  the  centre.  We 
should  have  20  barrels  in  each  row  (Fig.  91),  weighing  in  all  20.190  = 
3920  lbs.  These  could  not  well  be  placed  closer  than  3  feet  from 
end  to  end,  or,  say,  1307  lbs.,  per  running  foot  of  wall;  of  this  amount 
only  one-half  will  thrust  against  wall,  or,  say,  G50  pounds.  The  ra- 
dius of  the  barrel  is  about  20".      If  Figure  92  represents  three  of 

the  barrels,  and  we  make  A  B  = 
ti\  =  ^  the  load  of  the  flour  barrels, 
per  running  foot  of  walls,  it  is  evi- 
dent that  D  B  will  represent  the 
horizontal  thrust  on  wall,  per  run- 
ning foot.  As  D  B  is  the  radius,  and 
as  we  know  that  A  D  =  2  D  B  or  = 
'2  radii,  we  can  easily  find  A  B,  for : 


A  D2 


Fie.  92. 

DB2  — AB2 


DB2r=-^or 
3 


DBr=^^z= 


or  4.  D  B'^  —  D  B2  : 


:0,578.tt'.  •- 


/I 
/I 

\\ 

IHH 

I 
I 


^ O 


^3"     i,n 

Or,  A  =  0,578.  ?o,  (64) 

Where  h  =  the  horizontal  thrust,  in  lbs., 
against  each  running  foot  of  wall,  u!,=r  one- 
half  the  total  load,  in  lbs.,  of  barrels  coming 
on  one  foot  of  floor  in  width,  and  half  the 
span.  In  our  case  we  should  have  : 
74  =  0,578.650  =  375  lbs. 

Now,  to  find  the  height  at  which  this 
thrust  would  be  applied,  we  see,  from  Fig- 
ure 91,  that  at  point  1  the  thrust  would  be 
from  one  line  of  barrels ;  at  point  2,  from 
two  lines  ;  at  point  3,  fi-om  three  lines, 
etc. ;  therefore,  the  average  thrust  will  be 
at  the  centre  of  gravity  of  the  triangle 
A  B  C,  this  we  know  would  be  at  one-third  = 
the  height  A  B  from  its  base  A  C 

Now  B  C  is  eciual  to  6?-  or  six  radii  of    >-  »    > 

^  SCALE  OF  X'-'erGHrs.CLB5j 


i2 


24 36 


iCALE  OFL£AfOr/fJ  (/NCffEiJ 
O  ISOO     3O0O  45QO 


the  barrels  ;  further,  A  C  =  3r,  therefore  : 


ng.93. 


A  B2  =  3C.r2  — 9r2 


AB 


25.?-2,  and  A  B  ^  5?-;  therefore, -— =:  l|r. 


148 


SAFE    KLILDING. 


To  this  must  be  ailJetl  the  radius  A  D  (below  A)  so  tliat  the  central 
point  of  thrust,  O  in  this  ease,  would  be  above  the  beam  a  distance 
2/  =  2|.r. 

Where  y  =  the  height,  in  inches,  above  floor  at  which  the  average 
thrust  takes  place.     AVhere  r  ==  radius  of  barrels  in  inches. 

Our  radius  is  10",  therefore  : 

7/ =  202-" 

Xow,  in  Figure  93  let  A  B  C  D  be  the  12"  wall,  A  the  floor  level, 
G  M  the  central  axis  of  wall,  and  A  O  =  26|;  draw  O  G  horizon- 
tally ;  make  G  II  at  any  scale  equal  to  the  permanent  load  on  A  D, 
which,  in  this  case,  would  be  the  former  load  less  the  wind  and  snow 
allowances  on  wall  and  roof,  or 

3832  — (16. 30 4"  13.30)  =  2962,  or,  say  3000  lbs. 

Therefore,  make  G  II  =  3000  lbs.,  at  any  scale ;  draw  B.l  =  h  = 
375  lbs.,  at  same  scale,  and  draw  and  prolong  G  I  till  it  intersects 
D  A  at  K.     The  pressure  at  K  will  be  /;  =  G  I  =  3023. 

*We  find  the  distance  M  K  measures 
MK  =  x  =  3i". 

Thei'efore,  from  formula  (44)  the  stress  at  D  will  be  : 

3023   I    (3313023  _   I   5(3  ii3s_  ("oi,  compression) 
144  ~      12.144        n-  V  i-  y 

While  at  A  the  stress  would  be,  from  formula  (45)  : 

3023_g3|^3023__j^  ^^g_  ,      tension),  so  that 
144  12.144  ^ 

the  wall  would  be  safe. 

The  writer  has  given  this  example  so  fully  because,  in  a  recent 
case,  where  an  old  building  fell  in  New  York,  it  was 
claimed  that  the  walls  had  been  thrust  outwardly  by 
flour  barrels  piled  against  them. 

Narrow  Piers.  Where  piers  between  openings  are 
narrower  than  they  are  thick,  calculate  them,  as  for 
isolated  piers,  using  for  d  (in  place  of  thickness  of 
wall)  the  width  of  pier  between  openings ;  and  inP- 
place  of  L  the  height  of  opening.  The  load  on  the 
pier  will  consist,  besides  its  own  weight,  of  all  walls, 
girders,  floors,  etc.,  coming  on  the  wall  above,  from 
centre  to  centre  of  openings. 

Wind-pressure.        To  calculate  wind-pressure,  assume 
it  to  be  normal  to  the  wall,  then  if  A  B  C  D,  Figure    - 
94,  be  the  section  of  the  whole  wall  above  ground 
(there  being  no  beams  or  braces  against  wall). 


DMC    K 

Fig.  94. 


WIND-PKESSUKE.  14l> 

Make  O  D  =  ^.  A  I>  =  — ;  draw  G  II,  the  vertical  neutral  axis 

of  the  -whole  mass  of  wall,  make  G  II,  at  any  convenient  scale,  equal 
to  the  whole  weight  of  wall;  draw  II  I  horizontally  equal  to  the  total 
amount  of  wind-pressure. 

This  wind-pressure  on  vertical  surfaces  is  usually  assumed  as  being 
equal  to  00  lbs.  per  square  foot  of  the  surface,  provided  the  surface 
is  flat  and  normal,  that  is,  at  right  angles  to  the  wind.  If  the  wind 
strikes  the  surface  at  an  angle  of  45°  the  pressure  can  be  assumed 
at  15  lbs.  per  foot. 

It  will  readily  be  seen,  therefore,  that  the  gi-eatest  danger  from 
wind,  to  rectangular,  or  square  towers,  or  chimneys,  is  when  the 
wind  strikes  at  right  angles  to  the  widest  side,  and  not  at  right 
angles  to  the  diagonal.  In  the  latter  case  the  exposed  surface  is 
larger,  but  the  pressure  is  much  smaller,  and  then,  too,  the  resistance 
of  such  a  structure  diagonally  is  much  greater  than  directly  across 
its  smaller  side. 

In  circular  structures  multiply  the  average  outside  diameter  by 
the  height,  to  obtiin  the  area,  and  assume  the  pressure  at  15  lbs.  per 
square  foot.  In  the  examples  already  given  we  have  used  15  lbs., 
where  the  building  was  low,  or  where  the  allowance  was  made  on  all 
sides  at  once.  AVhere  the  wall  was  high  and  supposed  to  be  normal 
to  wind  we  used  30  lbs.  Referring  again  to  Figure  94,  continue  by 
drawing  G  I,  and  prolonging  it  till  it  intersects  D  C,  or  its  prolonga- 
tion at  K.  Use  formulae  (44)  and  (45)  to  calculate  the  actual  pres- 
sure on  the  wall  at  D  C,  remembering  that,  x  =  M  K ;  where  M 
the  centre  of  D  C,  also  that, 

p  =  G  I;  measured  at  same  scale  as  G  H. 
Remember  to  use  and  measure  everything  uniformly,  that  is,  all 
feet  and  tons,  or  else  all  inches  and  pounds.  The  wind-pressure  on 
an  isolated  chimney  or  tower  is  calculated  similarly,  except  that  the 
neutral  axis  is  central  between  the  walls,  instead  of  being  on  the 
wall  itself;  the  following  example  will  fully  illustrate  this. 

Example. 

Is   the   chimney,    Figure   87,   safe   against    wind- 
Wind- pressure 
on  Chimney,     jjressure? 

We  need  examine  the  joints  A  and  E  only,  for  if  these  are  safe 
the  intermediate  ones  certainly  will  be  safe  too,  where  the  thickening 
of  walls  is  so  symmetrical  as  it  is  here. 

The  load  on  A  we  know  is  47  tons,  while  that  on  E  is  057  tons. 


150  SAFE    BUILDING. 

Now  the  wind-pressure  down  to  A  is  : 

Pa  =  10.30.15  =  4500  lbs.,  or  =  2^  tons. 

On  base  joint  E,  the  wind-pressure  is  : 

P  =  r2f.l50.15  —  28500  lbs.,  or  =  U\  tons. 

We  can  readily  see  that  the  wind  can  have  no  appreciable  effect, 
but  continue  for  the  sake  of  illustration.  Draw  Pa  horizontally  at 
half  the  height  of  top  part  A  till  it  intersects  the  central  axis  G, ; 
make  G,  11,  at  any  convenient  scale  =  47  tons,  the  load  of  the  top 
part ;  draw  H,  I,  horizontally,  and  (at  same  scale)  =  2\  tons  =  the 
wind-pressure  on  top  part ;  draw  G  I, ;  then  will  this  represent  the 
total  pressure  (from  load  and  wind)  at  K,  on  joint  A  A,. 

TTmi  Formula  (44)  to  get  the  stress  at  A,  where  x  =.  K,  M,  =  9", 
or  4  feet;  and  p  =  G,  I„  which  we  find  scales  but  little  over  47 
tons ;  and  Formula  (45)  for  stress  at  A.  For  d  the  width  of  joint 
we  have  of  course  the  diameter  of  base,  or  lOf  feet.     Therefore 

Stress  at  A,  =  |^  +  6.  ^^  =  +  1,71  tons,  per  square 

foot,  or^-iIi^^^  =  -f-24  lbs.  (compression),  per  square  inch,  and 

14: -i 

Stress  at  A  =1^  —  6.  4^2  =  +  0,7    tons,    per    square    foot,    or 
0,7.2000  —  _j_^^  j|^g_  (compression),  per  square  inch. 

To  find  the  pressure  on  base  E  E. ;  draw  P  G  horizontally  at  half 
the  whole  height;  make  G  M  =  657  tons  (or  the  whole  load)i  and 
draw  U  K  horizontally,  and  ==  141  tons  (or  the  whole  wind-pressure). 
Draw  G  K  and  (if  necessary)  prolong  till  it  intersects  E  E,  at  K.  From 
formula  (44)  and  (45)  we  get  the  stresses  at  E,  and  E  :  p  being  =  G  K 
=  658  tons  ;  and  x  =  M  K  =  20",  or  If  feet.  For  d  the  width  of 
base  we  have  the  total  diameter,  or  16  feet.   Therefore  stress  at 

E  -  ^^  4-  6.  ^iHl  =  -4-  7  tons  per  sq.  ft.,  or  ^4^  =  +07    lbs. 
■^'"~151~      151.16         '  ^  I'l-i 

,  T       G58       p  658.1|_ 
(compression),  per  square  inch,  and  stress  at  t.  _  — —  o.~— — 

4-  1|  tons  (compression),  per  square  foot,  or    ^j-^     =  -f-  23  lbs. 

(compression)  per  square  inch. 

There  is,  therefore,  absolutely  no  danger  from  wind. 
Strength  of  Corbels  carrying  overhanging  parts  of  the  walls, 

Corbels,  etc.,  should  be  calculated  in  two  ways,  first,  to  see 
>  Scale  of  weights.  Fig.  87,  applies  to  G,  H,,  etc.,  but  not  to  G  IM. 


STRENGTH   OF   CORBELS. 


151 


wlicthcr  the  corbel  itself  is  strong  enough.  We  consider  the  corbel 
as  a  lever,  and  use  either  Formula  (25),  (2G)  or  (27) ;  according  to 
how  the  overhang  is  distributed  on  the  corbel,  usually  it  will  be  (25). 
Secondly,  to  avoid  crushing  the  wall  immediately  under  the  corbel, 
or  possible  tipping  of  the  wall.  "Where  there  is  danger  of  the  latter, 
long  iron  beams  or  stone-blocks  must  be  used  on  (op  of  the  back  or 
wall  side  of  corbel,  so  as  to  bring  the  weight  of  more  of  the  wall  to 
bear  on  the  back  of  corbel. 

To  avoid  the  former  (crushing  under  corbel)  find  the  neutral  axis 
G  H  of  the  whole  mass,  above  corbel.  Figure  95 ;  continue  G  II  till 
it  intersects  A  B  at  K,  and  use  Formulas  (44)  and  (45). 

If  M  be  the  centre  of  A  B,  then  use  a:  =  K  M,  and  p  =  weight  of 
corbel  and  mass  above ;  remembering  to  use  and  measure  all  parts 
alike ;  that  is,  either,  all  tons  and  feet,  or  all  pounds  and  inches. 

Example. 

Calculation  of  '^  brick  tower  has  pilasters  30"  wide,  projecting 
corbel.  jg".  Qf^  Q^e  side  the  toioer  is  engaged  and  fur  rea- 
sons of  planning  the  pilasters  cannot  be  carried  down,  but  must  he  sup- 
ported on  granite  corbels  at  the  main  roof  level,  which  is  36  feet  below 
top  of  tower.  The  loall  is  24"  under  corbel,  and  averages  from  there 
to  top  IG",  offsets  being  on  both  sides,  so  that  it  is  central  over  2i"  wall. 
What  thickness  should  the  granite  corbel  be  ? 
If  we  make  the  pilaster  hollow,  of  8"  walls,  we  will  save  weight  on 

the    corbel,    though 
w^e  lose  the  advant- 
[*  5"Cr  ■*!  age  of  bonding  into 

wall  all  way  up.  Still, 
we  will  make  it  hol- 
low. 

First,  find  the  dis- 
K  Vl    S  I  I  !  tancc  of  neutral  axis 

M-N  (Figure  9G), 
of  pilaster  from  24" 
wall,  which  will  give 
the  central  point  of 


H 


M^- 

-4:_ 

~'T' 

_N 

% 

Ro^, 

1 
.1. 

F 

ig.  96. 

We  use  rule  given  in  the  first  article, 


Fi2.  95 
application  of  load  on  corbel 
and  have 

^  _  8.30.12  +  8.8.4  +  8.8.4  ^    ,  „ 
8.30+8.8  +  8.8  ^ 

The  weight  of  pilaster  (overlooking  corbel)  will  be  GG24  lbs.,  or  3^  tons 


152 


SAFE   BUILDIXG. 


Assuming  that  only  30"  in  length  of  the  wall  will  come  on  back 
part  of  corbel,  the  weight  on  this  part  would  be  8640  lbs.,  or  4^  tons. 

The  bending  moment  of  6624  lbs.  at  9^"  from  support  on  a  lever 
is  (see  Formula  27)  : 

7)1  =  6624.9A-  =  61824  (lbs.  inch). 

From  Formula  (18)  we  know  that: 

771 


(7) 


From  Table  I,  Section  Number  3,  we  have: 
And  from  Table  V,  we  have  for  average  granite  : 

^  (7)="" 

Inserting  these  values  in  Formula  (18)  we  have: 
61824        ,   „         „       61824 


5.d^,  or  d^  : 


=  68,7,  thei-efore,  d 


180  '  900 

We  should  make  the  block  10"  deep,  however,  to  work  better  with 
brickwork.     This  would  give  at  the  wall  a  shearing  area  =  10.30  = 

300  square  inches,  or  — —  =  22  lbs.,  per  square  inch,  which  granite 

will  certainly  stand ;  still,  it  would  be  better  to  corbel  out  brickwork 
under  granite,  which  will  materially  stiffen      ,5^  4/^ 

Pressure  ^^^   strengthen   the    block.       Q  Q) 

under  corbel.     To  find  the  crushing  strain        !<■ S^AZ~~^ 

on  bi'ick  wall  under  corbel,  find  the  central 
axis  of  both  loads  by  same  rule  as  we  find 
centre  of  gravity ;  that  is,  its  distance  A  Kf 
from  rear  of  wall  will  be  (Figure  97) 
41.12  +  31.331 

^*  +  ^3 
or,  say,  22",  the  pressure  at  this  point  will, 


224" 


"SI 


\ 


IM 


•*"      22 


'A 


•24—^ 


Fig.  97. 

of  course,  be  =  4J  -j-  3J  =  7J  tons,  or  p  :^  15340  lbs. 

The  area  will  be  =  24.30  =  720  square  inches,  while  K  M  meas- 
ures 10"  ;  we  have,  then,  from  Formulte  (44)  and  (45)  stress  at 


B  = 


15340 


15340.10 


stress  at  A  = 


720 
15340 


720.24 
15340.10 


74  lbs.  (compression),  and 


31  lbs.  (tension). 


720  720.24 

There  would  seem  to  be,  therefore,  some  tendency  to  tipping,  still 


PRESSURE   UNDER   CORBELS.  153 

we  can  pass  it  as  safe,  particularly  as  much  more  than  30"  of  the  wall 
will  bear  on  the  rear  of  corbel. 

If  the  wind  could  play  against  inside  wall  of  tower,  it  might  help 
to  upset  the  corbel,  but  as  this  is  impossible,  its  only  effect  could  be 
against  the  pilaster,  which  would  materially  help  the  corbel  against 
tipping. 


CHAPTER  V. 


3p 


/ 


Fig.  98. 


HE  manner  of  laying  arches  has  been  de- 
scribed in  the  previous  chapter,  while  in  the 
first  chapter  was  given  the  theory  for  calcu- 
lating their  strength ;  all  that  will  be  necessary, 
therefore,  in  this  chapter  will  be  a  few  practical 
examples.  Before  giving  these,  however,  it 
will  be  of  great  assistance  if  we  first  explain 
the  method  of  obtaining  grapldcally  the  neu- 
tral axis  of  several  surfaces,  for  which  the 
arithmetical  method  has  already  been  given 
Neutral  axis  (P-  ^O-  To  find  the  vertical  neu- 
found  graphi-  tral  axis  of  two  plane  surfaces 
*^^"^'  A  B  E  F  and  B  C  D  E,  proceed 

as  follows :  Find  the  centres  of  gravity  G„  of 
the  former  surface  and  G,  of  the  latter  surface. 
Through  these  centres  draw  G„  II„  and  G,  H, 
vertically. 

Draw  a  vertical  line  a  c  anywhere,  and  make 
ab  at  any  scale  equal  to  surface  A  B  E  F,  and  at 
same  scale  make  6  c  =  B  C  D  E.  From  any 
point  o,  draw  o  a,  ob  and  oc.  From  the  point  of 
intersection  g,oi  oc  with  H,  G„  draw  ^r,  </„  paral- 
lel 0  b  till  it  intersects  II„  G„  at  ^„ ;  from  ^„  draw 
ffii  0  parallel  with  o  a  till  it  intersects  o  c  at  g ; 
through  g  draw  g  G  vertically  and  this  is  the  de- 
sired vertical  neutral  axis  of  the  whole  mass. 

Where  there  are  many  parts  the  same  method 
C  is  used. 

We  will  assume  a  segmental  arch  divided  into 
five  equal  parts.  Calling  part  A  B  C  D  =  No. 
I ;  part  D  C  E  F  =  No.  II,  etc.,  and  the  verti- 
cal neutral  axis  of  each  part,  No.  I,  No.  II,  etc. 


ARCHES. 


155 


Draw  a  f  vertically  and  at  any  convenient  scale,  make 
a6  =  No.  lor  ABCD    '  J'      D 

6c  =  No.norDCEF,  [HWWiWWX 

cd  —  No.  Ill 
de=^  No.  TV  and 
ef=  No.  V 
From  any  point  o  draw  oa,  oh,  oc,  od,  oe 
and  of. 

From  point  of  intersection  1  of  verti- 
cal axis  No.  I  and  oa,  draw  1  g  parallel 
with  0  b  till  it  intersects  vertical  axis 
No.  TI  at  g :  then  draw  g  h  parallel  Avith 
o  c  till  it  intersects  vertical  axis  No.  Ill 
at  h  ;  similarly  draw  li  i  parallel  with  o  d 
to  axis  No.  IV  and  i  J  parallel  with  o  e 
to  axis  No.  V,  and,  finally,  draw  J  5  par- 
allel with  0  /  till  it  intersects  a  o  (or 
its  prolongation)  at  5.  A  vertical  axis 
A 


Fig.  99. 


through  5  is  the  ver- 
tical neutral  axis  of 
the  whole  mass. 
Pi-olong  j  i  till  it 
intersects  o  a  at  4, 
i  h  till  it  intersects 
o  a  at  3,  and  h  g 
till  it  intersects  o  a 
at  2. 

A  vertical  axis 
through  4  will  then 
be  the  vertical 
neutral  axis  of  the 
mass  of  Nos.  I,  II, 
III  and  IV ;  a  ver- 
tical axis  through 
3  will  be  the  verti- 
cal neutral  axis  of 
the  mass  of  Nos.  I, 
Fig  100.  II   and    III;    and 

through  2  the  vertical  neutral  axis  of  the  mass  of  Nos.  I  and  II.     Of 
course  the  axis  throuirh  1  is  the  vertical  neutral  axis  of  No.  I. 


)00   loo    30e  -#00 


SCA,L£    Ofi    P£eT. 


We  will  now  take  a  few  practical  examples. 


156 


SAFE    BUILDING. 


Example  I. 

In  a  solid  brick  wall  an  opening  3  feet  wide  is  bricked  over  with 
an  8-inch  arch.     Is  this  stronrj  enoucjh  ?  ^ 

The  thickness  of  the  wall  or  arch,  of  course,  does  not  matter, 
where  the  wall  is  solid,  and  we  need  only  assume  the  wall  and  arch 
to  be  one  foot  thick.  If  the  wall  were  thicker,  the  arch  would  be 
correspondingly  thicker  and  stronger,  so  that  in  all  cases  where  a 
Arch  In  load  is   evenly    distributed    over  an  arch  we   will 

soil  wa  .  (.Qnsider  both  always  as  one  foot  thick.  If  a  wall  is 
hollow,  or  there  are  uneven  loads,  we  can  either  take  the  full  actual 
thickness  of  the  arch,  or  we  can  proportion  to  one  foot  of  thickness 
of  the  arch  its  proportionate  share  of  the  load. 

In  our  example  we  as- 
sume everything  as  one  foot 
thick.  The  load  coming  on 
the  half-arch  B  J  I  L  Fig. 
100,  will  be  enclosed  by  the 

Voussoirs  lines  A  L 

arbitrary.  ^^^^  I A  at 

60°  with  the  horizon.     We 

divide  the   arch  into,  say, 

four  equal  voussoirs  B  C  = 

CF  =  FG  =  GJ.    (The 

manner  of  dividing  might, 

of  course,  have  been  arbi- 
trary as  well  as  equal,  had 

we  preferred.)     Draw  the 

radiating   lines  through  C, 

F  and  G,  and   from  their 

upper  points  draw  the  ver-  

tical  lines  to  D,  Yu  and  11.  Fig.  loi. 

Now  find  the  weight  of  each  slice,  rememhering  always  to  include  the 

weight  cf  the  vouspoir  in  each  slice.     We  have,  then,  approximately, 
No.  I      (A  B  C  D)  =  3'    X    i'  X  112  pounds  =  1G8  pounds 
No.  II    (DCFE)  =  2i'x    i'xll2       "       =r  119       " 
No.  Ill  (E  F  G  H)  ==  U'  X  ^V  x  112       "       =    73       " 
No.  IV  (II  G  J  I)  =    I'  X  ^V  X  112       "       =    43       " 

Total  =  403       " 

1  For  convenience,  most  of  the  figures  are  duplicated  :  the  first,  showing  man- 
ner of  obtaining  the  horizontal  thrust ;  the  second,  the  manner  cf  obtaining  the 
line  of  pressure. 


HORIZONTAL    riJF.SSUUE.  157 

As  the  arch  is  eviilcntly  heavily  loaded  at  the  centre,  we  assume 
the  point   a  at  one-third   the  height  of   B  L  from  the  top,  or 

Jj  a  =z  -i-  =2i"  and  draw  the  horizontal  line  a  4. 
3  ^ 

As  previously  explained,  find  the  neutral  axes : 

„     ,       ^  .  1  '7i  of  part  No.  I, 

Horizontal  •^'       ^  ' 

pressures.      2  ^3  of  parts  No.  T  plus  No.  II, 

3  ffa  of  i)arts  Nos.  I,  II  and  III,  and 

4  ffi  of  the  whole  half  arch. 

Now  make  at  any  scale  1  7,  =  1G8  units  =  No.  I ;  similarly  at  same 

scale  2  g«=  168  -f-  110  units  =  287  =  No  I  and  No.  II,  and 

3^3=287+  73  units  =  360  =  No.  I  +  No.  II  +  No.  III. 

4  gi=  360  -\-  43  units  =  403  =  weight  of  half  arch  and  its  load. 

Now  make :  C  Zi  =  |  C  L,  =  2§". 

Similarly,  F  4  =  A  F  L„  =  21"";  also,  0^3  =  ^01.3  =  2§", 

And,  J  l^  =  iJI  =  2§". 

Through  g„  g«,  g^  and  g^  draw  horizontal  lines,  and  draw  the  lines 

1  Z„  2  L,  3  I3  and  4  U  till  they  intersect  the  horizontal  lines  at  /«„  /jo,  A, 

and  7^4 ;  then  will  g,  h,  measured  at  same  scale  as  1  g^  represent  the 

horizontal  thrust  of  A  B  C  D ;  ^^  Ju  the  horizontal  thrust  of  A  B  F  E  j 

g^  hi  the  horizontal  thrust  of  A  B  G  II  and  174  hi  the  horizontal  thrust 

of  the  half  arch  and  its  load.     In  this  case  it  happens  that  the  latter 

is  the  greatest,  so  that  we  select  it  as  our  horizontal  pressure,  and 

make  (in  Fig.  101)  a  0  =  ^4  Z!4  =;  620  pounds,  at  any  convenient  scale. 

Now  (in  Fig.  101)  make  o  &  =  168  pounds  =:  No.  I. 

6c  =  119       "       =No.  II., 

c  c?  =    73       "       =  No.  III. 

f/  e  =    43       "       =  No.  lY. 

Draw  oh,  oc,  od  and  oe. 

^  .  Now  begin  at  a,  draw  a  1  parallel  with  0  a  till  it  in- 

Curve  of  °    ^_       '  „  ,  „  , 

pressure,  tersects  axis  No.  1  at  1 ;  from  1  draw  1 1,,  parallel  with 

0  h  till  it  intersects  axis  No.  II  at  i, ;  from  /,  draw  «,  i«  parallel  with  0  c 

till  it  intersects  axis  No.  Ill  at  io;  from  /o  draw  L  i^  parallel  with  o  d 

till  it  intersects  axis  No.  IV  at  i^ ;  from  l^  draw  23  A''4  parallel  with  0  e. 

A  curve  through  the  points  a,  7v„  /iC.,  Kz  and  K^  (where  the  former 

lines  intersect  the  voussoir  joint  lines)  and  tangent  to  the  line  a  1  ?',  u  (3 

Ki  would  be  the  real  curve  of  pressure.     The  amount  of  pressure  on 

joint  C  L,  would  be  concentrated  at  7v,  and  would  be  equal  to  0  6 

(measured  at  same  scale  as  a  b,  etc.).     The  pressure  on  joint  F  Lo 

would  be  concentrated  at  Ko  and  be  equal  to  0  c.     The  pressure  on 


158 


SAFE    BUIT.niXG. 


joint  G  L3  would  be  concentrated  at  K^  and  equal  to  0  d.  The  pres- 
sure on  the  skew-back  joint  J  I  would  be  concentrated  at  Ki  and  be 
equal  to  0  e.  The  latter  joint  evidently  suffers  the  most,  for  not  only 
has  it  Rot  the  greatest  pressure  to  bear,  but  the  curve  of  pressure  is 
farther  from  the  centre  than  at  any  .other  joint.  We  need  calculate 
this  joint  only,  therefore,  for  if  it  is  safe,  the  others  certainly  are  so, 

too.     By  scale  we  find  that  J  /Ci  measures  2i-",  or  7^4 
Stress  at  ex-  •'  %       .  n  \  r 

trades  and      is  (x  =)  1^"  from  the  centre  of  joint;  we  find  fur- 
intrados.       ^.j^^,^.  ^-^^^^  ^  ^  scales  740  units,  therefore  (p  =)  0  e  ^= 
740  jrounds. 

The  width  of  joint  is,  of  course,  8",  and  the  area  =:  8  x  12  =  96 
square  inches ;  therefore,  from  Formula  (44) 


Stress  at  edge  J  =:-;^-^-j-6 


li_ 


16,26 


96     '         96.  8 
and  from  (45) 

Stress  at  ed2;e  I  = 6.  ——^ — i  =  —  0,85 

°  9G  96.  8 

Or  the  edge  J  would  be  subject  to  the  slight  compression  of  16^ 

pounds,  and  edge  at  I  to  a  tension  of  a  little  less  than  one  pound  per 

square  inch.     Tlie  arch,  therefore,  is  more  than  safe. 

Example  II. 
A  four-inch  rowlock  brick  arch  is  built  between  two  iron  beams,  of 
five  feet  span,  the  radius  of  arch  being  foe  feet.    The  arch  is  loaded  at 
the  rate  of  150  pounds  per  square  foot.     Is  it  safe  ? 

In  this  case  Ave  will  divide  the  top  of  arch  A  D  into  five  equal  sec- 
ions  and  assume  that  each  section  carries  75  pounds  —  (which,  of 

course,  is  not  quite  correct).    We  find  the  horizontal 

Brick  floor-    ^  ^^^.  ^         .      ,         ,  ,    r  -,  r-    ■, 

arch,  pressures  (I  ig.  102)  gJi^Qihc,  etc.,  as  betore,  and  find 


100  200  Joo  ^'oc  yoo  eoo 


JC^.LE  oF  peer. 


Fig.  102. 

again  g^  h  the  largest  and  equal  to  575  pounds.   We  now  make  (Fig. 
103)  at  any  convenient  scale,  0  a  z=  g-Jii  =  bl5  pounds,  and  ab-= 


BKICK    KLOOU-AIICII. 


159 


bc^cd  =  de=:  e  /■=  75  pounds  and  draw  o  a,  o  b,  o  c,  etc.   We 

now  find  the  broken  line  a  1  /,  i.,  i^  i^  Kr,  where : 

rt  1  is  parallel  v.'ith  oa  1  /,  is  parallel  with  o  h 

i,  h   "  "     o  c  ii  ii     "  "     od 

hU   "  "     oe  iJCi  «  "of 

In  this  case  again  evidently  the  greatest  stress  is  on  the  skew-back 

joint  C  D,  for  it  not  only  has  the  greatest  pressure  o  f,  but  the  curve  of 


Fig.   103. 

JCALe  op  F£BT. 

pressure  passes  farther  from  the  centre  than  at  any  other  joint.  We 
find  that  C  K^  scales  \\  inches,  therefore  the  distance  of  Kr,  from  the 
centre  is  (x  =)  |",  We  scale  o  f  and  find  it  scales  GOO  units,  or 
(jo  )=  690  pounds.  The  joint  is  4"  wide  and  its  area  =  4  x  12  =  48 
square  inches. 

From  Formulae  (44)  and  (45)  we  have  then : 

Stress  at  C  =  ■ 


G90   I    g  GOO.  f  _ 
•.«'"'"    ■  48.  4  ■ 

c.  *  -n       COO       „  600.  f 

Stress  at  D  =  — —  —  6.  -——^  ■ 
48  4b.  4 


-\-  30,6  pounds  and 
:  — 1,8  pounds. 


The  arch,  therefore,  is  perfectly  safe. 
Example  III. 
Two  iron  heams,  Jive  feet  apart,  same  as  before,  but  filed  with  a 
straight  7"  hollow  f  re-clay  arch.     The  load  per  foot  to  be  assumed  at 
140  pounds.   Is  the  arch  safe  ? 

Fireproof  Of  the  350  pounds  on  the  half  arch  we  will  assume 

floor-arch.  gQ  pounds  to  come  on  each  of  the  blocks  and  30 

j_ 


D 


t^'f 


h4..«:--r:.... 


hj-"--- 


:.C_-E..-..o.2ij....i 


.^di 


^3.. 


A 


TWi 


M  -    S3 
S5 


160 


SAFE    BL'ILDIXG. 


pounds  on  the  skew-back.  We  then  (in  Fig.  104)  find,  as  before  the 
horizontal  pressures,  g,  7i„  g.  lu,  etc.  Again  we  find  the  largest  pres- 
sure to  be  Qi  h,  and  as  it  scales  2040  units,  we  make  (in  Fig.  105)  at 
any  convenient  scale  and  place  o  a^  riJii=  2040  pounds.  We  also 
make  ah  =  hc  =  cd  =  (le^=?>0  units  and  e/=  30  units. 

Draw  oa,  oh,  oc,  etc.   Drawing  the  lines  parallel  thereto,  beginning 
at  a  we  get  the  line  a  1  t,  h  h  u  lU,  same  as  before    Imagining  a  joint 
-r 


2o1-o 


i\oo 


200  ^oo  ^oo  yoo  t 


OF    POUMDJ. 


SC/>^UE  op    FEBT. 


Fig.    105. 

at  C  D  this  would  evidently  be  the  joint  with  greatest  stress,  for  the 
same  reasons  mentioned  before.  We  find  C  K^  scales  2§",  and  as  C  D 
scales  7^"  the  point  K^  is  distant  from  the  centre  of  joint. 

(X  =  )3|-2|r=l" 

as  o/ scales  2100  units  or  pounds,  and  the  joint  is  7;i"  deep  with  area 

=  7^.  12  =  87  square  inches,  we  have : 

„^  ^  r,        2100    ,    „   2100.  1         ,    A,  ,.  1         A 

Stress  at  C  =  -—-  +  6.  ^    =-}-  44,14  pounds  and 

0/  o7.  i-^ 


Stress  at  D 


2100 


6. 


2100.1 


4,14  pounds. 


87.  7i 

The  arch,  therefore,  would  seem 
perfectly  safe.  But  the  blocks  are 
not  solid ;  let  us  assume  a  section 
tlirough  the  skew-back  joint  C  D 
to  be  as  per  Fig.  106.  We  should 
have  in  Formulte  (44)  and  (45) 
X,  p,  and  the  depth  of  joint  same 
as  before,  but  for  the  area  we 
should  use  a  ==  3.1^.12  =  54  square 
inches,  or  only  the  area  of  soUd 


parts  of  block.  Therefore  we  should  have : 


FimU'nOOF    FLOOK-AIICII. 


IGl 


Stress  at  C  = 


2100 


Stress  at  D  = 


54 
2100 


G. 


-j-  71  pounds,  and 


2100.  1. 
54.  7f 

21*^0.^=  _^  6,71  pounds. 


54  54.   7i 

There  need,  therefore,  be  no  doubt  about  the  safety  of  the  arch. 

Example  IV. 
Over  a  20-inch  brick  arch  of  8  feci  clear  span  is  a  centre  pier  16' 
wide,  carrying  some  ttco  tjns  tveif/ht.    On  each  side  of  pier  is  a  window 
opening  2^  feet  wide,  and  beyond,  piers  similarly  loaded.     Is  the  arch 

safe? 

.     ^  ,    ,  We  divide  the  half  arch  into  five  equal  voussoirs. 

Arch  In  front  t 

wal  I    concen-  The  amounts  and  neutral  axes  of  the  dillcrent  vous- 

trated  loads.  gQjj,g^  j^jjj  lyads  coming  over  each,  are  indicated  in 
circles  and  by  arrows;  thus,  on  the  top  voussoir  E  B  (Fig.  107)  we 
have  a  load  of  2100  pounds,  another  of  02  pounds,  and  the  weight  of 
voussoir  or  228  pounds.  The  neutral  axis  of  the  three  is  the  verti- 
cal through  G,  (Fig.  108).     Again  on  voussoir  E  F  (Fig.  107)  we 


UCAI-E   op    FEBT, 


Fig.    107. 

have  the  load  174  pounds,  and  weight  of  voussoir  228  pounds;  the 
vertical  neutral  axis  of  the  two  being  throujih  G„  (Fig.  108).  Simi- 
larly we  get  the  neutral  axes  G„„  G,v  and  Gy  (Fig.  108)  for  each  of 
the  other  voussoirs.     Now  remembering  that  1  r/,  (Fig.  107)  is  tho 


162 


SAFE   UUILUING. 


neutral  axis  of  and  equal  to  the  voussoir  B  E  and  its  load  ;  2  g^  the 
neutral  axis  of  and  cipal  to  the  sum  of  the  voussoirs  B  E  and  E  F 
and  their  loads,  etc.,  we  find  the  horizontal  thrusts  ^r,  /i„  g^  lu,  gj  hi, 
etc.  The  last  g^  Ih  is  again  the  largest,  and  we  find  it  scales  7850 
units  or  pounds. 

The  arch  being  heavily  loaded  we  selected  a  at  one-third  from  the 
top  of  A  B.  We  now  make  (Fig.  108)  a  o=  7850  pounds  or  units 
at  any  scale ;  and  at  same  scale  make  ab=  2390  pounds ;  b  c  =  402 
pounds;  c  d  =  432  pounds ;  d  e  =  2956  pounds,  and  e  f=  18G8 
pounds.  Draw  o  b,  o  c,  o  d,  etc.  Now  draw  as  before  a  1  parallel 
with  0  a  to  axis  G, ;  also  1  i,  parallel  with  o  &  to  axis  G„ ;  i,  h  parallel 
with  0  c  to  G,u,  etc.   We  then  again  have  the  points  a,  /f„  K«,  K^,  Kt 


O    too  2oo  5oo  4-00  ?oo  ftoo 
5CA1.E  OF    tOtJtiOf 


/C<ftUE    OF    FEET, 


Fig.     108. 

and  iTj  of  the  curve  of  pressure.  As  K^  is  the  point  farthest  from  the 
centre  of  arch-ring  and  at  the  same  time  sustains  the  greatest  pressure 
{of)  we  need  examine  but  the  joint  C  D ;  for  if  this  is  safe  so  are  the 
others.     We  insert,  then,  in  Formulaj  (44)  and  (45)  for 

p  =  of=  11250  pounds,  and  as  K^  C  measures  Gi", 
X  =  10"  —  Ci"  =  3i"  ;  also  as  the  joint  is  20"  wide, 
a  =  12.  20  =  240  square  inches. 


ARCH    WITH    AISUT.MENT. 


1G3 


Therefore 


Stress  at  C  = 


11250 


Stress  at  D  = 


240 

11250 

240 


The  arch  is,  therefore,  safe. 


^  11250.3.1    ,  ,,„     ,    , 
240.  2o"  ~  "^    pounds,  and 

.  11250. 3A     „     , 
o.  — -^  =  —  3  pounds. 

240.  20  '■ 


Example  V. 

A  12-inch  brick  semi-circular  arch  has  12  foot  span,  A  solid  brick 
wall  is  built  over  the  arch  to  a  level  with  one  foot  above  the  keystone. 

The  abutment  piers  are  5  feet  high  to  the  spring  of  arch  and  are  each 
Zfeel  wide,  including,  of  course,  the  tcidlh  nf  skew-backs.  Are  the  arch 
and  piers  safe  ? 

Arch  in  fence-       As  before,  we  will   assume   arch,  pier,  and  wall 

ment^  "  '  over  arch,  each  one  foot  thick.     "We  will  divide  the 

load   over  arch   into   seven   equally  wide   slices.     This  will  make 

uneven  voussoirs,  but 
this  does  not  matter, 
as  our  joint  lines  (and 
voussoirs)  are  only 
imaginary  anyhow, 
and  not  necessarily  of 
the  shape  of  the  ac- 
tual voussoirs,  which 
in  brick  Avould,  of 
course,  be  repre- 
sented by  each  single 
brick.  The  amount 
of  the  sums  of  each 
voussoir  and  its  load, 
and  the  vertical  neu- 
tral axes  of  the  differ- 
ent sets  are  given  by 
the  ai'rows  and  lines 
G„  Go,  Gs,  etc.  (in 
Fig.  110).  When 
considering  the 
safety  of  the  abut- 
ment we  treat  it  ex- 
actly the  same  as  the  voussoirs  (and  loads)  of  the  arch ;  that  is,  we 


1G4  SAFE    BU1LDI^^G. 

take  tlie  whole  weight  of  the  abutment,  viz.,  C  D  E  F I H  C  and  find 
its  neutral  a.>:is  Gg. 

Returning  now  to  the  arch,  we  go  through  the  same  iirocess  as 
before.  AVe  find  the  horizontal  pressu.es  (Fig.  109)  ff,  h„  g.^  h,  etc. 
In  this  case  we  find  that  the  last  pressure  f/;  1^  is  not  as  large  as  (j^  h ; 
therefore  we  adopt  the  latter;  it  scales  1425  units  or  pounds.  We 
now  make  (Fig.  110)  a  o  =^  1425  pounds  ;  and  a  6  =  251  pounds  ; 
6  c  r=  280  pounds  ;  c  (/  =  373  pounds,  etc. ;  g  h  is  equal  to  the  last 
section  of  arch  or  1 782  pounds.  We  continue,  however,  and  make 
7t  i  =  4G00  pounds  =  the  weight  of  abutment.  Draw  o  a,  o  b,  o  c, 
etc.,  to  0  i.  Then  get  the  tangents  to  the  curve  of  pressure,  as  be- 
fore, viz. :  a  1  /,  h  h  u  h  u  K^ ;  we  now  continue  i^  K^,  which  is  par- 
allel with  0  h  till  it  intersects  the  vertical  axis  G^  of  the  abutment  at  t, 
and  from  thence  we  draw  ?j  Kg  parallel  with  o  i. 

We  will  now  examine  the  base  joint  I  H  of  pier. 
"'"^''"I\)utment.  I  JQ  scales  10;^",  and  as  the  pier  is  3G"  wide,  Ka  is 
7|"  from  the  centre  of  joint.     The  area  is  a  =  12.36  =  432  and  the 
pressure  isp  =  o  i  =  0100  pounds. 
Therefore, 

Stress  at  I  =  ^^  +  6.  ^'^  =  +  48  pounds, 

and 

StressatH  =  ^^-6.|l|?fII  =  -G  pounds. 
432  432. oO 

There  is,  therefore,  a  slight  tendency  for  the  pier  to  revolve  around 
the  point  I,  raising  itself  at  H  ;  still  the  tendency  is  so  small,  only  6 
pounds  per  square  inch,  that  we  can  safely  pass  the  pier,  so  far  as 
dano-er  from  thrust  is  concerned. 

Joint  C  D  at  the  spring  of  the  arch  looks  rather  dangerous,  how- 
ever, as  lo  h  cuts  it  so  near  its  edge  D.  Let  us  examine  it.  D  A'; 
measures  If,  therefore  K^  is  U"  from  the  centre  of  joint,  which  is 
12"  wide.   The  area  is,  of  course,  a  =  12.  12  =  144  and  the  pressure 

J)  =  0  h  =  4G00  pounds.   Therefore, 

Stress  at  D  =  ^^  +  G.^^*  =  +  104  pounds, 


and 


Stress  at  C  =  ^  -  G.  ^  =  -40  pounds. 


144 

It  is  evident,  therefore,  that  the  arch  itself  is  not  safe,  and  it  should 
be  designed  deeper ;  that  is,  the  joints  should  be  made  deeper  (say, 
16"),  and  a  new  calculation  made. 


TIE-RODS    TO    AUCIIES. 


1G5 


Tie-rods  to  ^^'  instead  of  an  abutment-pier,  wo  had  used  an 

arches,  ji-on  tic-rod,  its  sectional  area  would  have  to  be  suf- 
ficient to  resist  a  tension  equal  to  the  greatest  horizontal  thrust  o  a; 


/o 

/o 

i  / 

•m 

!/■ 

/ 

■h 

; 

i    / 


!  / 


^  . 


Fig.  I  10. 


■JCAI-EOFFEET-  •• 

and  care  should  be  taken  to  proportion  the  washers  at  each  end,  large 


166  SAFE   BUILDING. 

enough  that  they  may  have  sufficient  bearing-surface  so  as  not  to 

crusli  the  material  of  the  skew-backs. 

Thus,  in  Example  III,  Fig.  105,  if  we  should  place  the  iron  tie-rods 

to  the  beams  5  feet  apart,  they  would  resist  a  tension  equal  to  five 

times  the  horizontal  thrust  o  a,  which,  of  course,  was  calculated  for 

1  foot  only,  or 

t  =  5.  2040  =  10200  pounds. 

The  safe  resistance  of  wrought-iron  to  tension  is  from  Table  IV,  12000 

pounds  per  square  inch ;  we  need,  therefore  : 

10200       „„.  .     , 

— — —  =  0,8o  square  inches 

of  area  in  the  rod ;  or  the  i-od  should  be  1  ^V"  diameter.  A  1"  or 
even  |"  rod  would  probably  be  strong  enough,  however,  as  such 
small  iron  is  apt  to  be  better  welded,  and,  consequently,  stronger, 
and  the  load  on  the  arch  would  probably  be  a  "  dead  "  one. 

As  the  end  of  rod  will  bear  directly  against  the  iron  beam,  the 
washers  need  have  but  about  ^"  bearing  all  around  the  end  of  the 
rod,  so  that  the  nut  would  probably  be  large  enough,  and  no  washer 
be  needed. 

Example    VI. 

A  pier  28"  ivide  and  10'  Jiir/h  supports  iv:o  abull'mg  semi-circular 
arches  ;  the  rifjht  one  a  20"  Irick  arch  of  8'  span ;  the  left  one  an  8" 
brick  arch  of  3'  span.  The  loads  on  the  arches  are  indicated  in  the 
Figure  111.     Is  the  pier  safe  f 

Uneven  arches  The  loads  are  so  heavy  compared  to  the  weight  of 
with  central  pier,  the  voussoirs,  that  we  will  neglect  the  latter,  in  this 
case,  and  consider  the  vertical  neutral  axis  of  and  the  amount  of 
each  load  as  covering  the  voussoirs  also ;  except  in  the  case  of  the 
lower  two  voussoirs,  where  the  axes  are  considerably  affected.  We 
find  the  curve  of  pressure  of  each  arch  as  before. 

For  the  large  arch  we  would  have  the  curve  through  a  and  i,  for 
the  small  one  through  a,  and  i, ;  the  points  i  and  z,  being  the  inter- 
sections of  the  curves  with  the  last  vertical  neutral  axis  of  each  arch. 

Now  from  i  draw  i  x  parallel  with  of,  and  from  z,  draw  i,  x  (back- 
wards), but  parallel  with  o^f  till  the  two  lines  intersect  at  x.  Now 
make  f  g  parallel  with  and  =  o,  f,  and  draw  g  h  vertically  =  2600 
pounds  z=  the  weight  of  the  pier  from  the  springing  line  to  the  base 
(1'  thick).  Draw  o  g  and  o  h.  Now  returning  to  x,  draw  x  y  paral- 
lel with  g  0  till  it  intersects  the  neutral  axis  of  the  pier  at  y,  and  from 
y  draw  y  z  parallel  with  o  h  till  it  intersects  the  base  joint  C  D  of 


THRUST   ON   CENTRAL   PIER. 


1G7 


pier  (or  its  prolongation)  at  K,.     Continue  also  x  y  till  it  intersects 
the  springing  joint  A  B  and  K.     Now,  then,  to  get  the  stresses  at 


Fig.   111. 


joint  A  B  we  know  that  the  width  of  joint  is  28",  therefore  a  =  28.12 
=  336  •  further  p  =  o  g-=  19250  pounds,  and  as  B  K  scales  one 
inch,  K  is  distant  13"  from  the  centre  of  joint,  therefore 


Stress  at  B  =  1?S^  +  6.  ^'^:^  =  -\-2ll  pounds. 


and 


336 


,   .        19250       p 
Stress  at  A  = —  6 


336.  28 


19250. 13 


■102  pounds. 


336         '     336.  28 

The  arch,  therefore,  cannot  safely  carry  such  heavy 

Thrust  on  ,,  ^  ^     c     i 

central  pier,  loads.     The  pier  we  shall  naturally  expect  to  lind 

still  more  unsafe,  and  in  effect  have,  remembering  that  joint  D  C  is 
28"  wide,  therefore,  area  336  square  inches,  and  as  K,  distant  54" 
from  centre  of  joint,  and  p  =  o  A  =  21 750  pounds. 

Stress  at  D  =  ^V  6.  ^S^  =  +  ^13  pounds. 


336 


336.  28 


168 


SAFE    BUILDING. 


and 


Stress  at  C  = 


21750 


21750.54 


=  —  684  pounds. 


336  336.  28 

Relief  through  The  construction,  therefore,  must  be  radically 
iron-work. changed,  if  the  loads  cannot  be  altered.  If  the 
arches  are  needed  as  ornamental  features,  they  should  be  constructed 
to  carry  their  own  weight  only,  and  iron-work  overhead  should  carry 
the  loads,  and  bear  either  nearer  to,  or  directly  over,  the  piers,  as 
farther  trials  and  calculations  might  call  for.  If  this  is  done  the  wall 
To  avoid  should  be  left  hollow  under  all  but  the  ends  of  iron- 

cracks,  work  until  it  gets  its  "  set " ;  that  is,  until  it  has 
taken  its  full  load  and  deflection ;  and  then  the  wall  should  be 
pointed  with  soft  "  putty  "  mortar. 

Example  VII. 
The  foundations  of  a  building  rest  on  brick  piers  6'  wide  and  18' 
apart.  The  piers  are  joined  by  32"  brick  inverted  arches  and  tied  to- 
gether 8'  above  the  spring  of  the  arch.  Piers  and  arches  are  3'  thick. 
Load  on  central  piers  is  72  tons  ;  on  end  pier  60  tons.  Is  this  con- 
struction safe  ? 

Inverted  We  will  first  examine  the  inner  or  left  pier.     The 

arches,  pjer  being  3'  thick  and  the  load  72  tons,  each  1'  of 

thickness  will,  of  course,  carry  -''^  =  24  tons.    The  width  from  centre 

to  centre  of  piers  is  just  24',  so  that  each  running  foot  of  wall  under 


^CAUB   «F  To 


J<AUE    oP   Fe 


Fig.  112. 

arch  will  receive  a  pressure  of  one  ton.    Now  all  we  need  to  do  is  to 
imagine  this  pressure  as  the  load  on  the  arch.    We  can  either  draw 


INVEUTKD   AKCIIES.  169 

the  aTcli  upside  down  with  a  load  of  one  ton  per  foot,  or  we  can  make 
the  drawing  with  the  arch  in  correct  position  and  the  weights  press- 
ing upward,  as  sliown  in  Fig.  1 1 2.    We  divide  the  load  into  five  equal 

,.  „,         slices,  each  ubout  2'  wide,  therefore  =  2  tons  each. 

Strength  or  '  ,  „  ^  r    ■> 

arch.  Wc  make  a  b  =  2  tons  ;  b  c  ^=  2  tons,  etc.,  and  nnd 

the  horizontal  pressures  </,  /<„  (/^  h,  etc.,  same  as  before.  Again,  g^  h 
is  the  largest,  measuring  13  units  or  tons.  Now  make  o  a  =  V3  tf.ns, 
and  draw  ob,o  c,  etc.  Construct  the  line  a  1  IC,  or  curve  of  pressure 
same  as  before.  Joint  L  M  is  evidently  the  most  strained  one.  "We 
find  INI  A'o  measures  11",  and  as  the  arch  is  32"  (=  M  L),  of  course, 
Ki  is  5"  from  the  centre  of  joint.  The  area  of  joint  is  a  =  32.  12  = 
384.  We  scale  of,  the  pressure  at  K^,  and  find  that  it  measures  16^ 
units,  or  33000  pounds,  therefore 

stress  atM  =  ?^»  +  6.i^^=+168  pounds 

stress  at  L  =  ^4^^-  G.  '^^  =  +  G  pounds. 

^     ^    ,  The  arch,  therefore,  is  perfectly  safe.     Of  course, 

Central  ^  .  •      i.i 

pier.  the  left  pier  is  safe,  for  being  an  inner  pier  the  re- 
sistance Ka  X  of  the  adjoining  arch  to  the  left  will  just  counter-bal- 
ance the  thrust  of  our  arch,  or  K„  x.  But  at  the  end  (right)  pier  this 
is  different.  We,  of  course,  proportion  the  length  of  foundation  A  C, 
to  get  same  pressure  as  on  rest  of  wall  under  arch.  The  end  pier  car- 
ries 60  tons,  or  Y  =  20  tons  per  foot  thick.  The  pressure  per  running 
foot  on  wall  under  arch  we  found  to  be  one  ton,  therefore  A  C  should 
Thrust  on  ^®  ^0  feet  long.    The  half-arch  will  take  the  pressure 

end  pier,  of  10  feet  (from  A  to  B)  or  10  tons,  and  the  balance 
(10  tons)  will  come  on  B  C.  This  will  act  through  its  central  axis 
G  II,  which,  at  the  arch  skew-back,  will  be  half-way  between  the  end 
of  arch  J  and  outside  pier-line  I  F.  This  will,  of  course,  deflect  some- 
what the  abutment  (or  last  pressure)  line  K^  li  of  the  arch.  At  any 
convenient  place  draw  a,/,  vertically  equal  10  tons,  and  a,  o,  horizon- 
tally equal  13  tons,  the  already  known  horizontal  pressure.  Then 
Kt  II  is,  of  course,  parallel  with  and  equal  o,/,.     Now  make/,  g,  = 

10  tons  and  draw  o,  g,.  Now  on  the  pier  draw  H  K  parallel  with  o,  g,; 

11  being  the  point  of  intersection  of  G  II  and  K^  11.  As  the  pier  is 
tied  back  8'  above  the  arch,  we  take  our  joint-line  at  E  F,  being  8' 
above  D.  We  find,  by  scale,  that  K  is  58"  from  the  centre  of  E  F, 
the  latter  being  72"  wide.  The  area  is  a  =  72.  12  =  864.  And  as 
0,^,  scales  24  units,  the  pressure  is,  of  course,  24  tons  or  48000  pounds, 


170  SAFE    BUILDING 

therefore  * 

'  stress  at  F=.lg^+6.  "''''• ''=^  +  315  pounds. 
8G4      '  «G4.   72        ^  ^ 

and 

^      48000       „  48000.58 
Stress  at  E  =   ^^    —  6.    ^„^    ^^  =  —  20 /  pounds. 

Use  of  There  is,  therefore,  no  doubt  of  the  insecurity  of 

buttress,  the  end  pier.  Two  courses  for  safety  ?.re  now  open. 
Either  we  can  build  a  buttress  sufficiently  heavy  to  resist  the  thrust 
of  the  end  arch,  or  we  can  tie  the  pier  back.  The  former  case  is 
easily  calculated ;  we  simply  include  the  mass  of  the  buttress  in  the 
resistance  and  shift  the  axis  G  II  to  the  centre  of  gravity  of  the  area 
of  i^ier  and  buttress  up  to  joint  E  F.  Of  course,  the  buttress  should 
be  carried  up  to  the  joint-line  E  F,  but  it  can  taper  away  from  there. 
ygg  Qf  If  we  tie  back  with  iron  we  need  sufficient  area  to 

tie-rods,  resist  a  tension  equal  to  the  horizontal  pressure  a  o, 
which  in  this  case  is  13  tons  or  26000  pounds  per  foot  thick  of  waU. 
As  the  wall  is  3'  tliick,  the  total  horizontal  thrust  is  3.  26000  =  78000 
pounds.     The  tensional  stress  of  wrought-iron  being  12000  pounds 

per  square  inch,  we  need  =  6^  square  inches  area,  or,  say,  two 

wrought-iron  straps,  4"  x  |",  one  each  side  of  pier.  By  this  method 
the  inner  or  left  arch  becomes  practically  the  end  arch.  For  the  last 
two  piers  and  the  right  arch  become  one  solid  mass ;  and  not  only  is 
their  entire  weight  thrown  against  the  second  or  inner  arch,  but  the 
centre  of  gravity  of  the  whole  mass  shifts  to  the  centre  line  of  end 
arch,  or  in  our  case  0'  inside  of  the  end  pier ;  so  that  there  is  no  pos- 
sible doubt  of  the  strength  of  the  abutment.  There  is  one  element 
of  weakness,  however,  in  the  small  bearing  the  pier  has  on  the  skew- 
back  of  the  arch,  the  danger  being  of  the  pier  cracking  upwards  and 
settling  past  and  under  the  arch.  This  can  be  avoided,  as  already 
explained,  by  building  a  large  bond-stone  across  the  entire  pier,  form- 
ing skew-backs  for  the  arch  to  bear  against. 
Example  VIIL 
A  semi-circular  dome,  circular  in  plan,  is  40'  inside  diameter.  The 
shell,  is  5'  (hick  at  the  spring  and  2'  at  the  crown.  The  dome  is  of  cut- 
stone.     Will  it  stand  ? 

Calculation  ^^'^  ^^^"^  ^  section  (Fig.  1 14)  of  the  half-dome  and 

of  dome,  treat  it  exactly  the  same  as  any  other  arch.    The  only 

difference  is  in  the  assumption  of  the  weights.     Instead  of  assuming 

the  arch  1'  thick,  we  take  with  each  voussoir  its  entire  weight  around 


CIRCULAR   DOME. 


171 


one-lialf  of  the  dome.     Thus,  in  our  case,  wc  divide  the  section  into 
six  imaginary  voussoirs.    The  weight  on  each  voussoir  will  act  through 

l'?."...its  centre  of  grav- 
'  ity.      Now   weight 
No.  [  Avill  be  equal 
to  the  area  of  the  top 
voussoir  multiplied 
by    the    cii-cumfer- 
ence     of    a     semi- 
circle with  A  B  as 
radius.      Similarly, 
No.  II  will  be  equal 
to  the  area  of   the 
second     voussoir 
multiplied    by    tlie 
circumference  of  a 
semi-circle  with  A  C 
as  radius;  No.  Ill 
will  be  equal  to  the 
area  of  voussoir  3, 
multiplied    by    the 
circumference  of  a 
semi-circle  with    A 
D    as   radius,    etc. 
The   vertical    neu- 
tral axes  Nos.  I,  II, 
III,    etc.,    act,     of 
course,  through  the 
centres  of  gravity  of 


h«- 


'^6 


Fig.  113. 


their  respective  voussoirs.  Now  the  top  voussoir  measures  5'  G" 
i,y  an  average  thickness  of  2' 1"  or  5^^.2^2  =  1 1>46  or,  say,  lU  square 
feet.  As  A  B  (the  radius)  measures  2'  9",  the  circumference  of 
its  semi-circle  would  be  8',G.  Taking  the  weight  of  the  stone  at 
IGO  pounds  per  cubic  foot,  we  should  then  have  the  weight  of  No. 
I  =  11, .5.  8,  G.  IGO  =  15824  pounds,  or,  say,  8  tons. 
Similarly  we  should  have : 

No.    II  =  12,4.  25,1.  IGO  =    49798  pounds,  or,  say,    25  tons. 
No.  Ill  =  14,2.  40.       1G0=    90880        "  "  45     " 

No.  IV  =  17,4.  52,7.  1G0  =  14G71G        '=  "  73     " 

No.    V  =  21,2.  62,8.  IGO  =  213017        "  "         lOG     " 

No.  VI  =  28,7.   69,1.  IGO  =  317307        "  "         153     " 


172 


SAFE   BUILDING. 


We  now  make  a  b  =  8  tons,  &  c  =  25  tons,  c  d  ^  io  tons,  d  e  = 
73  tons,  e/=  106  tons  andfg  =  158  tons.  We  find  the  horizontal 
pressures  g,  h„  g^  ho,  etc.  (in  Fig.  113),  same  as  before.     In  this  case 


N^VIN'VNMV     mi"        "c" 


B     A 


70  o 


I  ' 


Fig.    114. 

•u-e  find  that  the  largest  pressure  is  not  the  last  one,  but  g^  Jh  which 
measures  78  units;  we  therefore  select  the  latter  and  (in  Fig.  114) 
make  a  o  =  g-^  h^  z=  78  tons.  Draw  o  b,  o  c,  o  d,  etc.,  and  construct 
the  line  a  1  i,  i^  is  U  h  IQi  same  as  before.  In  this  case  we  cannot 
tell  at  a  glance  which  is  the  most  strained  voussoir  joint,  for  at  joint 
U  I  the  pressure  is  not  very  great,  but  tlie  line  is  farthest  from  the 
centre  of  joint.     Again,  while  joint  J  L  has  not  so  much  pressure  as 


STRESS   AT    .JOINTS.  173 

the  bottom  joint  M  N,  the  line  is  farther  from  the  centre.   We  must, 
therefore,  e>:amine  all  three  joints. 

Wo  will  take  II  I  first.  The  width  of  joint  is  2'  5"  or  29".  The 
pressure  is  o  c,  which  scales  88  tons  or  17G000  pounds.  The  distance 
of  K«  from  the  centre  of  joint  P  is  10".  The  area  of  the  joint  is,  of 
course,  the  full  area  of  the  joint  around  one-half  of  the  dome,  or  equal 
to  II  I  multiplied  by  the  circumference  of  a  semi-circle  with  the  dis- 
tance of  P  from  a  </  as  radius.  The  latter  is  10'  G",  therefore  area 
=  2j^j.  33  =  80  square  feet  or  11520  square  inches,  therefore: 


c,           .  T       17G000   , 

Stress  at  I  = — -  4 

11520  ^ 

g  17G000.  16_   I 
11520.  20        "1 

-Co, 9 

o.            xTr       17GO0O 
Stress  at  11  =: —  - 

_(j   17G000.  1G__ 

-.35.? 

11520  11520.  29 

For  joint  J  L  we  should  have  :  the  width  of  joint  =:  50".  The 
pressure  =  o /:=  272  tons  or  544000  pounds.  The  distance  of  K^ 
from  the  centre  of  joint  is  8",  while  for  the  area  we  have  50.  CC§.  12 
=  40000  square  inches,  therefore : 


and 


o.  .  -r       544000   ,    „  540000. 8        y   ^n  o  i 

Stress  at  J  =:  — — —  4-  6.  — ttttt^ — ^  =4-  2G,G  pounds, 
40000    '        40000. 50        '        '    i'  ' 

o,  .  T        544000       „   540000.  8  ,    „  .  , 

Stress  at  L  = 6. =  4-  0,5  pound. 

40000  40000.  50       ~    '    ^ 


For  the  bottom  joint  M  X  we  should  have  the  width  of  joint  =  GO". 
The  pressure  =.  o  cj  =■  428  tons  or  85G000  pounds.  The  distance  of 
K^  from  the  centre  of  joint  is  4",  while  for  the  area  we  have  GO.  70|. 
12  =  50880  square  inches,  therefore  : 

,,       856000   .        85G000. 4 
Stress  at  M  =  -^^-g-  -f  G.  ^^gyo.  GO  =  +  "^'^  P°"°<^'' 
and 

^^       856000  856000.4 

Stress  at  Is  =  ^g^^  -  6.  ^^^^^^  ^^  =  + 10  pounds. 

The  arch,  therefore,  would  seem  perfectly  safe  except  at  the  joint 
H  I,  where  there  is  a  tendency  of  the  joint  to  open  at  H.  Had  we, 
however,  remembered  that  this  is  an  arch,  lightly  loaded,  and  started 
our  line  at  the  lower  third  of  the  crown  joint,  instead  of  at  the  upper 
third,  the  line  would  have  been  quite  different  and  undoubtedly  safe. 

The  dangerous  point  K«  in  that  case  would  be  much  nearer  the 
centre  of  joint,  while  the  other  lines  and  joints  would  not  vary  enough 
to  call  for  a  new  calculation,  and  we  can  safely  pass  the  arch.  One 
thing  must  be  noted,  however,  in  making  the  new  figure,  and  that  is 
that  the  horizontal  pressures  g^  h^,  goh^,  etc.  (in  Fig.  113),  would  have 


174  SAFE    RUXLDING. 

to  be  changed,  too,  and  would  boconie  somewhat  larger  than  before, 
as  the  line  a,  G  wouW  now  drop  to  the  level  of  a.,.  A  trial  will  show 
that  the  largest  would  again  be  ^j  k-^,  and  would  scale  80  units  or 
tons,  which  should  be  used  (in  Fig.  114)  in  place  of  a  o. 

-£— Thrust  of  -^^   regards  abutments,    there  usually 

dome,  are  none  in  a  dome ;  it  becomes  neces- 
sary, therefore,  to  take  up  the  horizontal  thrust,  either 
by  metal  bands  around  the  entire  dome,  or  by  dovetailing 
the  joints  of  each  horizontal  course.  We  must,  of  course, 
take  the  horizontal  thrust  existing  at  each  joint.  Thus,  if 
we  were  considering  the  second  joint  II  I  the  horizontal 
thrust  would  be  g.,  lu;  or,  if  we  were  considering  the  fifth  joint 
J  L  the  horizontal  thrust  would  be  g,,  lu^.  For  the  lower  joint  M  N 
we  might  take  its  own  horizontal  thrust  g^  h^  which  is  smaller  than 
^5  /«5,  provided  we  take  care  of  the  joint  J  L  separately ;  if  not,  we 
should  take  the  larger  thrust. 

Metal  bands.  If  we  use  a  metal  band  its  area  manifestly  should 
be  strong  enough  to  resist  one-half  the  thrust,  as  there  will  be  a  sec- 
tion in  tension  at  each  end  of  the  semi-circle,  or 

"=%)  (-) 

Where  a  =  area,  in  square  inches,  of  metal  bands  around  domes, 
at  any  joint. 

Where  h  =  the  horizontal  thrust  at  joint,  in  pounds. 

Where  (  —  j  =  the  safe  resistance  of  the  metal  to  tension,  per 

square  inch. 

In  our  case,  then,  for  the  two  lower  joints,  if  the  bands  are  wrought- 
iron,  we  should  have,  as  h  ■=  80  tons  =  160000  pounds. 
160000    _ 
"■~2^1'2000~~^ 
Or  we  would  use  a  band,  say,  5"  x  1^". 
Strength  of  ^^  dovetailed  dowels  of  stone  are  used,  as  shown  in 

dowels.  Y[g,  115,  there  should  be  one,  of  course,  in  every  ver- 
tical joint.  The  dowels  should  be  large  enough  not  to  tear  apart  at 
A  D,  nor  to  shear  off  at  A  E,  nor  to  crush  A  B.  Similarly,  care 
should  be  taken  that  the  area  of  C  G  -|-  B  F  is  sufllcient  to  resist  the 
tension,  and  of  II  C  to  resist  the  shearing. 

The  strain  on  A  D  will,  of  course,  be  tension  and  equal  to  one-half 
the  horizontal  thrust ;  the  same  formula  holds  good,  therefore,  as  for 


TIIKUST    OK    DOME.  175 

metal  bands,  excejjt,  of  course,  that  we  use  for  (yj  its  value  for  what- 
ever stone  we  select.  Supposing  the  dome  to  be  built  of  average  mar- 
ble, we  should  have  from  'J'able  V, 

( -T.  )  ==  70,  therefore, 

IGOOOO 

2.    lO 

Now  if  the  dome  is  built  in  courses  2'  G"  or  30"  high,  the  width  of 

dovetail  at  its  neck  A  D  would  need  to  be : 

1143 
A  D  =  -OQ-  =  38  inches. 

As  this  would  evidently  not  lave  sufficient  area  at  G  C  and  B  F 
we  must  make  A  D  smaller,  and  shall  be  compelled  to  use  either  a 
metal  dowel,  or  some  stronger  stone.  By  reference  to  Table  V  we 
find  that  for  bluestone, 


{lr^  =  UO,  therefore: 


/ 

160000       _, 

571 
A  D  =  -^  =19  inches. 

This  would  almost  do  for  the  lower  joint,  which  is  60"  wide,  for  if  we 
made  :  G  C  +  B  F  =  38"  and 

A  D  =  19"  it  would  leave 
60  —  38  +  19  =  3"  or,  say,  U"  splay ;  that  is, 
D  II  =:  1|".     Had  we  used  iron,  or  even  slate,  however,  there  would 
be  no  trouble. 

The  shearing  strain  on  either  A  E  or  H  C  will,  of  course,  be  etpial 
to  one-quarter  of  the  horizontal  thrust  or 
h 


bz= 


(66) 


Where  b  =  the  width  of  one-half  the  dowel,  in  inches  (A  E) 
Where  h  =  the  horizontal  thrust  at  the  joint,  in  pounds 
Where  d  =  the  height  of  the  course,  in  inches. 

Where  r-^j=  the  safe  resistance  to  shearing,  per  square  inch, 
of  either  dowel  or  stone  voussoir  (whichever  is  weaker). 

AVhen  the  shearing  stress  of  a  stone  is  not  known,  we  can  take  tlie 
tension  instead.     Thus  in  our  case  as  the  marble  is  the  weaker,  we 


176 


SAFE    BUILDING. 


should  use 


(  -^  1  =  70,  therefore 
IGOOOO 


A  E  =  H  C  = 


:=  1 9  inches. 


4.  30.   70" 

The  compression  on  A  B  need  not  be  figured,  of  course,  for  while 
the  strain  is  the  same  as  on  A  E,  the  area  of  A  B  is  somewhat  larger, 
and  all  stones  resist  compression  better  than  shearing. 
Cambered  When  a  plank  is  "cambered,"  sprung  up,  and  the 

plank  arch,  ends  securely  confirmed,  it  becomes  much  stronger 
transversely  than  when  lying  flat.     The  reason  is  very  simple,  as  it 
now  acts  as  an  arch,  and  forces  the  abutments  to  do  part  of  its  work. 
Such  a  plank  can  be  calculated  the  same  as  any  other  arch. 
Spanish  tile  Quite  a  curiosity  in  construction,  somewhat  in  the 

arches,  above  line,  has  been  recently  introduced  in  New  York 
by  a  Spanish  architect.   He  builds  floor-arches  but  3"  thick,  of  3  suc- 

'cessive  layers  of  1  "-thick  tiles,  up  to 
20'  span,  and  more.  His  arches  have 
withstood  safely  test-loads  of  700 
pounds  a  square  foot.  The  secret  of 
the  strength  of  his  arches  consists  in 
their  following  closely  the  curve  of 
pressure,  thus  avoiding  tension  in 
the  voussoirs,  as  far  as  possible.  But 
even  were  this  to  exist,  it  could  not 
open  a  joint  without  bodily  tearing  off 
several  tiles  and  opening  many  joints, 
as  shown  in  Fig.  IIG,  owing  to  the 
fact  that  each  course  is  thoroughly  bonded  and  breaks  joints  with  the 
course  below ;  besides  this,  each  upper  layer  is  attached  to  its  lower 
layer  by  Portland-cement  mortar.  Specimens  of  these  tiles  have  been 
tested  for  the  writer  and  were  found  to  be  as  follows : 
Compression  (12  days  old)  c  =  2D11  pounds 

or  say  (—.)z=  300  pounds. 

Tension  (10  days  old)  t  =  287  pounds 

or  say  (  -^  j  =  30  pounds. 

The  shearing  stress  of  the  mortar-joints  is  evidently  greater  than 
the  tension,  as  samples  tested  tore  across  the  tile  and  could  not  be 
sheared  off. 

The  modulus  of  rupture  (about  10  days  old)  was 


TILE    FLOOR    ARCH. 


177 


k=  91  pounds 

or,  say,  ^  -^  j  =  10  pounds. 

Of  course  older  specimens  would  prove  much  stronger- 
There  is  only  one  valid  objection  that  the  writer  has  heard  so  far 
against  these  arches,  and  this  is,  that  in  case  of  any  uneven  settle- 
ment they  might  prove  dangerous ;  as,  of  course,  the  margin  in  which 
the  curve  of  pressure  can  safely  shift  in  case  of  changed  surround- 
ings is  very  small.  The  writer  does  not  think  the  objection  very 
great,  however,  as  settlements  are  apt  to  ruin  any  arch  and  should 
be  carefully  provided  against  in  every  case.  The  arches  have  some 
very  great  advantages.  The  principal  one,  of  course,  is  their  light- 
ness of  construction  and  saving  of  weight  on  the  floors,  walls  and 
foundations.  Then,  too,  in  most  cases  iron  beams  can  be  entirely 
dispensed  with,  the  arches  resting  directly  on  the  brick  walls ;  of 
course,  there  must  be  weight  enough  on  the  wall  to  resist  the  hori- 
zontal thrust,  or  else  iron  tie-rods  must  be  resorted  to.  The  former 
is  calculated  as  already  explained  for  retaining  walls.  If  tie-rods 
are  used,  they  are  calculated  as  explained  above  for  the  example  of 
the  7"  flat  floor  arch.  An  example  of  these  3"-tile  arches  may  be  of 
interest. 

»  Example  IX. 

A  segmental  3"  tile  arch,  built  as  explained  above,  has  20'  clear  span 
with  a  rise  at  the  centre  of  20".  The  jloor  is  loaded  at  the  rate  of  150 
pounds  per  square  foot.     Is  the  arch  safp  ? 

Exam  pie  of  ^^^  divide  the  arch  into  five  parts  or  voussoirs, 

tile  arch,  each  voussoir  carrying  2'  of  floor  =  300  pounds ;  we 

make  (Fig.  118)  a&  =  300  pounds,  be  =  300  pounds,  etc.,  and  find 


h?-' 


aoeo  asoo  3eo« 


Fig     1 1  7. 
the  horizontal  pressures  (Fig.  117)  ^,  /«„  g^  h.,  etc.    The  last  one  g^h-, 
is  the  largest  and  scales  4500  units  or  pounds.     We  now  make  (Fig. 


178 


SAFE   BUILDING. 


118)  a  0  =  goh  =  4500  pounds  and  draw  ob,  oc,  od,  etc.  We  next 
construct  the  curve  of  pressure  a  K  and  find  that  it  coincides  as 
closely  as  possible  with  the  centre  line  of  the  arch.    This  means  that 


i^ 


Fig.    I  I  8. 

the  pressure  on  each  joint  will  be  uniformly  distributed.  That  on 
the  lower  joint  is,  of  course,  the  largest  and  is  =  o  /,  which  scales 
4700  pounds.  The  area  of  the  joint  is,  of  course,  a  =  3.  12  =  36 
square  inches,  therefore  the  greatest  stress  per  square  inch  will  be 

— ^  =  130  pounds  compression. 
36 

As  the  tests  gave  us  300  pounds  compression,  per  square  inch,  as 
safe  stress  of  a  sample  only  twelve  days  old,  the  arch  is,  of  course, 
perfectly  safe. 

If,  however,  instead  of  a  uniform  load,  we  had  to  provide  for  a 
very  heavy  concentrated  load,  or  heavy-moving  loads,  or  vibrations, 
it  would  not  be  advisable  to  use  these  arches. 

r  of  ^°  ^^''  ^^  l-iOVQ  simply  considered  the  danger  of 

sliding,  compression  or  tension  at  the  joints  of  an  arch; 
there  is,  however,  another  element  of  danger,  though  one  that  does 
not  arise  frequently,  viz. :  the  danger  of  one  voussoir  sliding  past  the 
other.  Where  strong  and  quick-setting  cements  are  used  this  danger 
is,  of  course,  not  very  great.  But  in  other  cases,  and  particularly  in 
large  arches,  it  must  be  guarded  against.  The  angle  of  friction  of 
brick  against  brick  (or  stone  against  stone)  being  generally  assumed 
at  30°,  care  must  be  taken  that  the  angle  formed  by  the  curve  of 
pressure  at  the  joint,  with  a  normal  to  the  joint  (at  the  point  of  in- 
tersection K)  does  not  form  an  angle  greater  than  30°.  If  the  angle 
be  greater  than  30°  there  is  danger  of  sliding ;  if  smaller,  there  is,  of 
course,  no  danger.  Thus,  if  in  Fig.  114  we  erect  through  Ki  a  nor- 
mal Ki  X  to  the  joint,  the  angle  A'  K^  i^  should  not  exceed  30°. 


Danger  of 


In  arches  with  heavv-concentratcd  loads  at  single 


shearing,  points,  thore  might,  in  rare  cases,  be  danger  of  the 


DKPTII    AT    CKOWN.  179 

load  shearing  right  througli  tlie  arch.  The  resistance  to  shearing 
would,  of  course,  be  directly  as  the  vertical  area  of  cross-section  of 
the  arch,  and  in  such  cases  this  area  must  be  made  largo  enough  to 
resist  any  tendency  to  shear. 

Depth  at  Arches  are  frequently  built  shallower  at  the  crown 

crown,  and  increasing  gradually'  in  depth  towards  the  spring, 
the  amount  being  regulated,  of  course,  by  the  curve  of  pressure  and 
Formula;  (44)  and  (45). 

To  establish  the  first  experimental  thickness  at  the  crown  of  an 
arch,  manv  engineers  use  the  empvrical  formula : 

Where  x  =  the  depth  of  arch,  at  crown,  in  inches. 

Where  r  =  the  radius  at  crown,  in  inches. 

Where  y  =  2l  constant,  as  follows  : 

For  cui  stone,  in  blocks  :  y  =  o,3 

For  brickwork  y  z=:  o,4 

For  rubble  work  ?/  =  o,45 

When  Portland  cement  is  used,  a  somewhat  lower  value  may  be 

assumed  for  y.     The  depth  thus  established  for  crown  is  only  exper- 

mental,  of  course,  and  must  be  varied  by  calculation  of  curve  of 

jjressure,  etc. 

Approximate  ^^  cases  where  the  architect  does  not  feel  the  ne- 

rule.  cessity  for  such  a  close  calculation  of  the  arch,  it 

will  be  feufficient  to  find  the  curve  of  pressure.     If  this  curve  of 

pressure  comes  within  the  inner  third  of  arch-ring,  at  every  point, 

the  arch  is  safe,  provided  the  thrust  on  each  joint,  divided  by  the 

area  of  joint,  does  not  exceed  one-half  oi  the  safe  compressive  stress 

of  the  material,  or : 

^-1  (±.\  (^^) 

a         2'\f  J 

Where  p  =  the  thrust  on  joint,  in  pounds. 

Where  a  =  the  area  of  joint,  in  square  inches. 

W^here  (  -^  ]=:  the  safe  resistance  to  compression,  per  square  inch, 

of  the  material. 

Vaulted  and  Vacdted  and  groined  arches  are  calculated  on  the 
Groined  Arches,  sj^^ie  princijiles  as  ordinary  arches  and  domes ; 
though,  in  groined  arches,  as  a  rule,  the  ribs  do  all  the  work,  the 
spandrels  between  the  ribs  being  filled  with  stone  slabs  or  other  light 
material ;  in  some  European  examples  even  flower-pots,  plastered  on 
the  under  side,  have  been  used  for  this  purpose. 


180 


CHAPTER  VL 


FLOOR   BEAMS   AND   GIRDERS- 


n  SLICE 

i  . 

(n-l)5LICE 

o 



4^  SLICE 

(t, , 

3^^      " 

5  St         .. 
M   _   .  —  -  _ 

(J, , 

X 

Fig.    I  I  9. 


Moment  ^^^^ 
of  Inertia,    -writer 

has  so  often  been 
asked  for  more  in- 
formation as  to 
the  meaning  of 
the  term  Moment 
of  Inertia  that  a 
few  more  words 
on  this  subject 
may  not  be  out  of 
place. 

All  matter,  if 
once  set  in  mo- 
tion, will  continue 
in  motion  unless 
— 1>^,  stopped  by  grav- 
ity, resistance  of 
the   atmosphere, 


N 


friction  or  some  other  force ;  similarly,  matter,  if  once  at  rest,  will 
so  remain  unless  started  into  motion  by  some  external  force. 
Formerly  it  was  believed,  however,  that  all  matter  had  a  certain  re- 
pugnance to  being  moved,  which  had  to  be  first  overcome,  before 
a  body  could  be  moved.  Probably  in  connection  with  some  such 
theory  the  term  arose. 

In  reality  matter  is  perfectly  indifferent  whether  it  be  in  motion 
or  in  a  state  of  rest,  and  this  indifference  is  termed  "  Inertia."  As 
used  to-day,  however,  the  term  Moment  of  Inertia  is  simply  a  symbol 
or  name  for  a  certain  part  of  the  formula  by  which  is  calculated  the 
force  necessary  to  move  a  body  around  a  certain  axis  with  a  given 


MOMKXT    OF    INKUTIA.  181 

velocity  in  a  certain  space  of  time ;  or,  what  amounts  to  the  same 
thing,  the  resistance  necessary  to  stop  a  body  so  moving. 

In  making  the  above  calculation  the  "  sum  of  the  product  of  the 
weight  of  each  particle  of  the  body  into  the  s(iuare  of  its  distance 
from  the  axis  "  has  to  be  taken  into  consideration,  and  is  part  of  the 
formula ;  and,  as  this  sum  will,  of  course,  vary  as  the  size  of  the 
body  varies,  or  as  the  location  or  direction  of  the  axis  varies,  it 
would  be  difficult  to  express  it  so  as  to  cover  every  case,  and  there- 
fore it  is  called  the  "  Moment  of  Inertia."  Hence  the  general  law 
or  formula  given  covers  every  case,  as  it  contains  the  ^Moment  of 
Inertia,  which  varies,  and  has  to  be  calculated  for  each  case  from  the 
known  size  and  weight  of  the  body  and  the  location  and  direction  of 
the  axis. 

In  plane  figures,  which,  of  course,  have  no  thickness  or  weight,  the 
area  of  each  ])article  is  taken  in  place  of  its  weight ;  hence  in  all 
plane  figures  the  Moment  of  Inertia  is  equal  to  the  "  sum  of  the  prod- 
ucts of  the  area  of  each  particle  of  the  figure  multiplied  by  the 
square  of  its  distance  from  the  axis." 

Calculation  of  Thus  if  we  had  arectangular  figure  (HO)  finches 
"^"tVa."*  °^  '"'  "wide  and  d  inches  deep  revolving  around  an  axis 
M-N,  we  would  divide  it  into  many  thin  slices  of  equal  height,  say 
n  slices  each  of  a  height  =  2.  X. 

The  distance  of  the  centre  of  gravity  of  the  first  slice  from  the 
axis  M-N  will,  of  course  be  =  ^.  2.  X.  =  1.  X 

The  distance  of  the  centi'e  of  gi-avity  of  the  second  slice  will  be  = 
3.x, 

that  of  the  third  slice  will  be=  5.  X, 
that  of  the  fourth  slice  will  be=  7.  X, 
that  of  the  last  slice  but  one  will  be  =  (2  n —  3).  X. 
and  that  of  the  last  slice  will  be  i=  (2  n  —  1).  X 

The  area  of  each  slice  will,  of  course,  be  =  2.  X.  6  ;  therefore  the 
Moment  of  Inertia  of  the  whole  section  around  the  axis  M-N  will 
be  (see  p.  8), 

i  =  2.  X.  h.  (1.  X)2  +  2.  X.  h.  (;3.  X)2-|-  2.  X.  h.  (5.  X)2  + 

2.  X.  h.  (7.  X)--f  etc +  2.  X.  6.  [(2;i  — 3).Xp 

-f  2.  X.  ?;.[(2n  — l).X]-or, 

i=2.X.3Z<.[l^+3-2 +  52+72  + etc +  (2;i  — 3)2 

+  (2u-l)2] 
now  the  larger  n  is,   that  is  the  thinner  we  make  our  slices,  the 
nearer  will  the  above  approximate : 


182 


SAFE    BUILDIKG. 


t  =  2.  X8.  6, 


[t--] 


=  8.  X3.  nS 


Therefore,  as  :  2.  X.  n  =  rf  we  have,  by  cubing, 

8.  X^.  n^  =  d^;  inserting  this  in  above,  we  have  ; 


J3. 


h.       h. 
—  or   - 


The  same  value  as  given  for  i  in  Table  I,  section  No.  29.  Of 
course  it  would  be  very  tedious  to  calculate  the  Moment  of  Inertia  in 
every  case  ;  besides,  unless  the  slices  were  assumed  to  be  very  thin, 
the  result  would  be  inaccurate;  the  writer  has  therefore  given  in 
Table  I,  the  exact  Moments  of  Inertia  of  every  section  likely  to  arise 
in  practice. 

The  Moment  of  Inertia  applies  to  the  whole  .<ec- 
Moment  of  c  t-.     •  .   , 

Resistance.    \\()X[,  the  "  Moment  of  Kesistance,    however,  apphes 

only  to  each  individual  fibre,  and  varies  for  each ;  it  being  equal  to 
the  Moment  of  Inertia  of  the  whole  section  divided  by  the  distance 
of  the  fibre  from  the  axis. 

Now  to  show  the  connec- 
tion of  the  ]\Ioments  of  In- 
ertia and  Resistance  with 
transverse  strains,  let  us 
consider  the  effect  of  a 
weight  on  a  beam  (sup- 
ported at  both  ends). 

If  we  consider  the  beam 
as  cut  in  two  and  hinged  at 


? 


I 


Fig.    120. 


the  point  A  (where  the  weight  is  applied).  Fig.  120;  further,  if  we 

consider  a  piece  of  rubber  nailed  to  the  bottom  of  each  side  of  the 

beam,  it  is  evident  that  the  O) 

effect  of  the  weight  will  be, 

as  per  Fig.  121. 

■•«     *  „*  i^^rt  Examin- 

Effect  of  load 

on  beam,  jj^g      ^^^^g 

closer  we  find  that  the  cor 

ners  of  the  beams  above  A 

(or  their  fibres)  wifl  crush 

each  other,  while  those  below  A,  are  separated  farther  from  each 

other,  and  the  piece  of  rubber  at  B  greatly  stretched.     It  is  evident, 

therefore,  that  the  fibres  nearest  A  experience  the  least  change,  and 


EFFECT   OF    LOAD. 


183 


that  the  amount  of  change  of  all  the  fibres  is  directly  proportionate  to 
their  distance  from  A  (as  the  length  of  all  Unes  drawn  parallel  to  the 
base  of  a  triangle,  are  proportionate  to  their  distance  from  the  axis) ; 
further,  that  the  fibres  at  A  experience  no  change  whatever.  Now, 
if  instead  of  considering  the  effect  of  a  load  on  a  hinged  beam  we 


-Y 


D 


Fig.  122. 

took  an  unbroken  beam,  the  effect  would  be  similar,  but,  instead  of 
beino-  concentrated  at  one  point,  it  would  be  distributed  along  the 
entire  beam;  thus  the  beam  A  B  D  C  (Fig.  122)  which  is  not  loaded, 
beco:i:es  when  loaded,  the  slightly  curved  beam  (A  B  D  C)  Fig.  123. 

?It  is  evident 
D  that  the  fibres 

-•^  '  along  the  upper 
edge  are  com- 
pressed or  A  B 
is  shorter  than 
before;  on  the 
other  hand  the 
Fig.  123.  fibres  along  CD 

are  elongated,  or  in  tension,  and  C  D  is  longer  than  before ;  if  we 
now  take  any  other  layer  of  fibres  as  E  F,  they— being  below  the 
neutral  (and  central)^  axis  X-Y  — are  evidently  elongated;  but  not 
so  much  so,  as  C  D  :  and  a  little  thought  will  clearly  show  that  their 
elongation  is  proportioned  to  the  elongation  of  the  fibres  C  D,  dir- 
ectly as  their  respective  distances  from  the  neutral  axis  X-Y.  It 
is  further  evident  that  the  neutral  axis  X-Y  is  the  same  length  as 
before,  or  its  fibres  are  not  strained;  it  is,  therefore,  at  this  point 
that  the  strain  changes  from  one  of  tension  to  one  of  compres- 
sion. 

In  Fi"-.  124  we  have  an  isometrical  view  of  a  loaded  beam. 


1  As  a  rule  the  neutral  axis  can  be  safely  assumed  to  be  central  but  it  is  not 
necessarily  so.  In  materials,  such  as  cast-iron,  stone,  etc.,  where  the  resistance 
of  the  fibres  to  comiircss ion  ami  tension  varies  greatly,  ihe  axis  will  be  far  from 
the  centre,  nearer  the  weaker  fibres. 


184 


SAFE   BUILDIXG. 


Rotation  around     ^-"^^  "^  °°^  consider  an  infinitesimally  thin  (cross) 

neutral  axis,    section  of  fibres  A  B  C  D  in  reference  to  their  own 

neutral  axis  M-N.    It  is  evident  that  if  we  wore  to  double  the  load 

on  the  beam,  so  as 
to  bend  it  still  more, 
that  the  fibres  along 
A  B  would  be  com- 
pressed towards  or 
would  move  towards 
the  centre  of  the 
^'S'  '  24.  beam  ;   the   fibres  a- 

long  D  C  on  the  contrary  would  be  elongated  or  would  move  away 
from  the  centre  of  the  beam. 

The  fibres  along  M-X,  being  neither  stretched  nor  compressed, 
would  remain  stationary. 

The  fibres  between  M-N  and  A  B  would  all  move  towards  the  cen- 
tre of  the  beam,  the  amount  of  motion  being  proportionate  to  their 
distance  from  M-X ;  the  fibres  between  M-N  and  D  C  on  the  contrary 
would  move  away  from  the  centre  of  the  beam  the  amount  of  motion 
being  proportionate  to  their  distance  from  M-X ;  a  little  thought, 
therefore,  shows  clearly  that  the  section  A  B  C  D  turns  or  rotates  on 
its  neutral  axis  M-I^,  whenever  additional  weight  is  imposed  on  the 
beam. 

This  is  why  we  consider  in  the  calculations  the  moment  of  Inertia 
or  the  moment  of  resistance  of  a  cross-section  as  rotating  on  its 
neutral  axis. 

Now  let  us  take  the  additional  weight  off  the  beam  and  it  will 
spring  back  to  its  former  shape,  and,  of  course,  the  fibres  of  the  in- 
finitesimally thin  section  A  B  C  D  will  resume  their  normal  shape ; 
that  is,  those  that  were  compressed  will  stretch  themselves  again, 
while  those  that  were  stretched  will   compress  themselves  back  to 

their  former  shape 
and  position,  and  those 
along  the  neutral  axis 
will  remain  constant ; 
or,  in  other  words, 
this  thin  layer  of  fibres 
A  B  C  D  can  be  con- 
sidered as  a  double 
wedge  -  shaped  figure 
Fig.  125.  A  B  A,  B.  M  N  D  C 


UESISTANCE   OF   FIBRES.  185 

D,  C,  (Fig.  125)  the  base  of  the  wedges  becoming  larger  or  smaller 
as  tlie  weight  on  the  beam  is  varied. 

Resistance  of  ^ow  to  proceed  to  the  calculation  of  the  resistance 
Wedge,  of  this  wedge.  It  is  evident  that  whatever  may  be 
the  external  strain  on  the  beam  at  the  section  A  B  C  D,  the  beam 
will  owe  wliatever  resistance  it  has  at  that  point  to  the  resistances 
of  the  fibres  of  the  section  or  wedge  to  compression  and  tension. 

Now  considering  the  right-hand  side  of  the  beam  as  rigid,  and  the 
section  A  B  C  D  as  the  point  of  fulcrum  of  the  external  forces,  we 
have  only  one  external  force  p,  tending  to  turn  the  left-hand  side  of 
the  beam  upwards  around  the  section  A  B  C  D,  its  total  tendency, 
effect  or  moment  m  at  A  B  C  D,  we  know  is  m=p.  x  (law  of  the 
lever). 

Now  to  resist  this  we  have  the  opposition  of  the  fibres  in  the 
•wedge  A  B  A,  B,  M  N  to  compression  and  the  opposition  to  tension 
of  the  fibres  in  the  wedge  D  C  D,  C,  M  N.  For  tlie  sake  of  conven- 
ience, we  will  still  consider  these  wedges,  as  wedges  but  so  infinites- 
imally  thin  that  we  can  safely  put  down  the  amount  of  their  con- 
tents as  equal  to  the  area  of  their  sides,  so  that  —  if  A  B  =  i  (the 
width  of  beam)  and  A  D  =  cZ  (the  depth  of  beam)  —  we  can  safely 

call  each  wedge  as  equal  to  6.  —  . 

Now  as  the  centre  of  gravity  of  a  wedge  is  at  ^  of  the  height  from 
its  base,  or  §  of  the  height  from  its  apex  (and  as  the  height  of  each 

wedo-e  is  =  —  Vt  would  be  =  1 .  —  =  4  ^ ^om  axis  ]\I  -  N.     The 
°  2  /  3      2         3 

moment  of  a  wedge  at  any  axis  M-N  is  equal  to  the  contents  of  the 
wedge  multiplied  by  the  distance  of  its  centre  of  gravity  from  the 
axis,  the  Avhole  multiplied  by  the  stress  of  the  fibres,  (that  is  their 
resistance  to  tension  or  compression).     Now  the  contents  of  each 

wedge  being  =  h.  ~,  the  distance  of  centre  of  gravity  from  M-N  =  — -j 

and  the  stress  being  say  =  s,  we  have  for  the  resistance  of  eaclt 
wedge 

=.  b.  —  .  —  .s 
2      3 

=  —'' 
Now  if  the  stress  on  the  fibres  along  the  extreme  upper  or  lower 
ed"-es  =  A;  (or  the  modulus  of  rupture),  it  is  evident  that  the  average 

°  k  k 

stress  on  the  fibres  in  either  wed2;e  will  =z  — - ,  or  s  =  ■— -  (for  the 


18G  SAFE    BUILDING. 

stress  on  each  fibre  being  directly  proportionate  to  the  distance  from 
the  neutral  axis  the  stress  on  the  average  will  be  equal  to  half  that 
on  the  base).     Now  inserting  —  for  s  in  the  above  formula,  and 

multiplying  also  by  2,  (as  there  are  two  wedges  resisting),  we  have 
the  total  resistance  to  rupture  or  bending  of  the  section  A  B  C  D 
(A.  B.  C.  D.) 

_b.d^    k     „ 

g--    ic 

b,cl2 

Xow,  by  reference  to  Table  I,  section  No.  2,  we  find  that  — g-  = 
Moment  of  Resistance  for  the  section  A  B  C  D  ;  therefore,  we  have 
proved  the  rule,  that  when  the  beam  is  at  the  point  of  rupture  at  any 
point  of  its  length  the  bending  moment  at  that  point  is  equal  to  the 
moment  of  resistance  of  its  cross-section  at  said  point  multiplied  by 
the  modulus  of  rupture. 

Where  girders  or  beams  are  of  wood,  it  becomes  of  the  highest 
importance  that  they  should  be  sound  and  perfectly  dry.  The  for- 
mer that  they  may  have  sufficient  strength,  the  latter  that  they  may 
resist  decay  for  the  longest  period  possible. 

Every    architect,   therefore,    should    study   thor- 
Formation  of  •'  ',,.., 

Wood,  oughly  the  different  kinds  of  timber  m  use  in  his 

locality,  so  as  to  be  able  to  distinguish  their  different  qualities.  The 
strength  of  wood  depends,  as  we  know,  on  the  resistance  of  its  fibres 
to  separation.  It  stands  to  reason  that  the  young  or  newly  formed 
parts  of  a  tree  will  offer  less  resistance  than  the  older  or  more  thor- 
oughly set  parts.  The  formation  of  wood  in  trees  is  in  circular  lay- 
ers, around  the  entire  tree,  just  inside  of  the  bark.  As  a  rule  one 
layer  of  wood  is  formed  every  year,  and  these  layers  are  known,  there- 
fore, as  the  "  annular  rings,*'  which  can  be  distinctly  seen  when  the 
trunk  is  sawed  across.  These  rings  are  formed  by  the  (returning) 
s;ap,  which,  in  the  spring,  flows  upwards  between  the  bark  and  wood, 
supplies  the  leaves,  and  returning  in  the  fall  is  arrested  in  its  altered 
state,  between  the  bark  and  last  annular  ring  of  wood.  Here  it  hard- 
ens, forming  the  new  annular  ring.  As  subsequent  rings  form 
around  it,  their  tendency  in  hardening  is  to  shrink  or  compress  and 
harden  still  more  the  inner  rings,  which  hardening  (by  compression) 
is  also  assisted  by  the  shrinkage  of  the  bark.  In  a  sound  tree,  there- 
fore, the  strongest  wood  is  at  the  heart  or  centre  of  growth.     The 


FOUMATXOX    or    WOOD. 


187 


heart,  however,  is  rarely  at  the  exact  centre  of  the  trunk,  as  the  sap 
flows  more  freely  on  the  side  exposed  to  the  effects  of  the  sun  and 
wind ;  and,  of  course,  the  rings  on  this  side  are  thicker,  thus  leaving 
the  heart  constantly,  relatively,  nearer  to  the  unexposed  side. 
Heart-Wood.  From  the  above  it  will  be  readily  seen  that  timber 

should  be  selected  from  the  region  of  the  heart,  or  it  should  be  what 
is  known  as  "  heart-wood."  The  outer  layers  should  be  rejected,  as 
they  are  not  only  softer  and  weaker,  but,  being  full  of  sap,  are  Uable 
to  rapid  decay.  To  tell  whether  or  no  the  timber  is  "  heart-wood  " 
one  need  but  look  at  the  end,  and  see  whether  it  contains  the  centre 
of  the  rin<Ts.  No  bark  should  be  allowed  on  timber,  for  not  only  has 
it  no  strength  itself,  but  the  more  recent  annular  rings  near  it,  are 
about  as  valueless. 

Medullary  Rays.  In  some  timbers,  notably  oak,  distinct  rays  are 
noticed,  crossing  the  annular  rings  and  radiating  from  the  centre. 
These  are  the  "medullary  rays,"  and  are  elements  of  weakness. 
Care  should  be  taken  that  they  do  not  cross  the  end  of  the  timber 
horizontally,  as  shown  at  A  in  Fig.  12G,  but  as  near  vertically  as 
possible,  see  B  in  Fig.  127.  The  beautiful  appearance  of  quartered 
oak  and  other  woods  is  obtained  by  cutting  the  planks  so  that  their 
surfaces  will  show  slanting  cuts  through  these  medullary  rays. 
Seasoning  ^^^  timber  cracks  more  or  less  in  seasoning,  nor 

cracks,  neo  j  these  cracks  cause  much  worry,  unless  they  are 
very  deep  and  long.  They  are,  to  a  certain  extent,  signs  of  the 
amount  of  seasoning  the  timber  has  had.  They  should  be  avoided, 
as  much  as  possible,  near  the  centre  of  the  timber,  if  regularly 
loaded,  or  near  the  point  of  greatest  bending  moment,  where  the 


m 


Fig.  12  6. 


Fig.  127. 


Fig.  128. 


Fig.  129. 


Fig.  130.       Fig.  13  1. 


loads  are  irregular.     If  timber  without  serious  cracks  cannot  be  ob- 
tained, allowance  should  be  made  for  these,  by  increasing  its  size. 

Vertical,  or  nearly  vertical  cracks  (as  C,  Fig.  128)  are  not  objec- 
tionable, and  do  not  weaken  the  timber.  But  horizontal  cracks  (as 
D,  Fig.  129),  are  decidedly  so,  and  should  not  be  allowed. 


188  SAFE    BUILDING. 

Knots.  Knots  ia  timber  are  another  element  of  weakness. 

They  are  the  hearts,  where  branches  grow  out  of  the  trunk.  If  they 
are  of  nearly  the  same  color  as  the  wood,  and  their  rings  gradually 
die  out  into  it,  they  need  not  be  seriously  feared.  If,  however,  they 
are  very  dark  or  black,  they  are  sure  to  shrink  and  fall  out  in  time, 
leaving,  of  course,  a  hole  and  weakness  at  that  place.  Dead  knott;,  — 
that  is,  loose  knots,  —  in  a  piece  of  timber,  mean,  as  a  rule,  that  the 
heart  is  decaying.  Knots  should  be  avoided  at  the  centre  of  a  beam, 
regularly  loaded,  and  at  the  point  of  greatest  bending  moment,  where 
the  loads  are  irregular.  The  farther  the  knots  (and  cracks)  are 
from  these  points  the  better. 

Wind-shakes.  Timber  with    "  wind-shakes  "  should  be  entirely 

avoided,  as  it  has  no  strength.  These  are  caused  by  the  wind  shak- 
ing tall  trees,  loosening  the  rings  from  each  other,  so  that  when  the 
timber  is  sawed,  the  wood  is  full  of  small,  almost  separate  pieces  or 
splinters  at  these  points. 

A  timber  with  wind-shakes  should  be  condemned  as  unsound. 

A  timber  with  the  rings  at  the  end  showing  nearly  vertical  (E 

Fig.  130)  will  be  much  stronger  than  one  showing  them  nearly  hori- 

zontaL     (F  Fig.  131.) 

To  tell  sound  timber,  Lord  Bacon  recommended 

Signs  of  sound  '     .       ,    .  ,  -j       th 

Timber,  ^g  speak  through  it  to  a  friend  from  end  to  end.     it 

the  voice  is  distinctly  heard  at  the  other  end  it  is  sound.  If  the 
voice  comes  abruptly  or  indistinctly  it  is  knotty,  imperfect  at  the 
heart,  or  decayed.  More  recent  authorities  recommend  listening  to 
the  ticking  of  a  watch  at  the  other  end,  or  the  scratching  of  a  pin 
on  its  surface.  If,  in  sawing  across  a  piece  it  makes  a  clean  cut,  it 
is  neither  too  green  nor  decayed.  The  same  if  the  section  looks 
bright  and  smells  sweet.  If  the  section  is  soft  or  splinters  up  badly 
it  is  decayed.  If  it  wets  the  saw  it  is  full  of  sap  and  green.  If  a 
blow  on  timber  rings  out  clearly  it  is  sound  ;  if  it  sounds  soft,  subdued, 
or  dull,  it  is  very  green  or  else  decayed.  The  color  at  freshly-sawed 
spots  should  be  uniform  throughout ;  timbers  of  darker  cross-section 
are  generally  stronger  than  those  of  lighter  color  (of  the  same  kind 
of  wood.) 

The  annular  rings  should  be  perfectly  regular.  The  closer  they 
are,  the  stronger  the  wood.  Their  direction  should  be  parallel  to 
the  axis  throughout  the  length  of  the  timber,  or  it  will  surely  twist 
in  time,  and  is,  besides,  much  weaker.  Where  the  rings  at  both 
ends  are  not  in  the  same  direction  the  timber  has  either  twisted  in 


SEASONING     TIMBF.R.  189 

growing,  or  has  a  "wandering  heart,"  —  that  is,  a  crooked  one. 
Such  timber  sliould  he  eondenined.  Besides  looking  at  the  rings  at 
the  end,  a  longitudinal  cut  near  the  heart  will  show  whether  it  has 
grown  regularly  and  straight,  or  whether  it  has  twisted  or  wandered. 

The  weight  of  timber  is  important  in  judging  its  quality.  I£  spec- 
imens of  a  wood  arc  much  heavier  than  the  well-known  weight  of 
that  wood,  when  seasoned,  they  may  be  condemned  as  green  and  full 
of  sap.  If  they  are  much  lighter  than  thoroughly  seasoned  speci- 
mens of  the  same  wood,  they  are  very  probably  decayed. 
Methods  of  Tredgold  claims  that  timber  is  "seasoned  "  when 

Seasoning.  j[.  jias  lost  one-fifth  of  its  original  weight  (when 
green) ;  and  "  dry  "  when  it  has  lost  one-third.  Some  timbers,  how- 
ever, lose  nearly  one-half  of  their  original  -woight  in  drying.  ]\Iany 
methods  are  used  to  season  or  dry  timber  quickly. 

The  best  method,  however,  is  to  stack  the  timber  on  dry  ground 
(in  as  dry  an  atmosphere  as  possible)  and  in  such  a  position  that  the 
air  can  circulate,  as  freely  as  possible  around  each  piece.  Sheds  are 
built  over  the  timber  to  protect  it  from  the  sun,  rain,  and  also  from 
severe  winds  as  far  as  possible. 

Timber  dried  slowly,  in  this  manner,  is  the  best.  It  will  crack 
somewhat,  but  not  so  much  so  as  hastily  dried  timber.  Many  proc- 
esses are  used  to  keep  it  from  cracking,  the  most  effective  being  to 
bore  the  timber  from  end  to  end,  at  the  centre,  where  the  loss  of 
material  does  not  weaken  it  much,  while  the  hole  greatly  relieves  the 
strain  from  shrinkage.  Some  authorities  claim  that  two  years'  ex- 
posure is  sufficient,  though  formerly  timber  was  kept  very  much 
longer.  But  even  two  years  is  rarely  granted  with  our  modern  con- 
ditions, and  most  of  the  seasoning  is  done  after  the  timber  is  in  the 
building.  Hence  its  frequent  decay.  There  are  many  artificial 
methods  for  drying  timber,  but  they  are  expensive.  The  best  known 
is  to  place  it  in  a  kiln  and  force  a  rapid  current  of  heated  air  past  it, 
this  is  known  as  "  kiln-drying."  It  is  very  apt  to  badly  "  check  "  or 
crack  the  wood.  To  preserve  timber,  besides  charring,  the  "  creo- 
soting  "  process  is  most  effective.  The  timber  is  placed  in  an  iron 
chamber,  from  which  the  air  is  exhausted ;  after  which  creosote  is 
forced  in  under  a  high  pressure,  filling,  of  course,  all  the  pores  which 
have  been  forced  open  by  the  suction  of  the  departing  air.  Creo- 
soted  wood,  however,  cannot  be  used  in  dwellings,  as  the  least  appli- 
cation of  warmed  air  to  it,  causes  a  strong  odor,  and  would  render 
the  buildin";  untenantable. 


190 


SATE    EUILDIXG. 


Manner  of 
Shrintca 


Decay  of 

Timber. 


In  slirinking,  the  distance  hclween  rings  remains 
"  constant,  and  it  is  for  this  reason  that  the  finest 
floors  are  made  from  quartered  stuff ;  for   (besides   their  greater 
beauty),  the  rings  being  all  on  end,  no  horizontal  shrinkage  will  take 
place ;     the     width    of 
boards     remaining    con- 
stant, and  the   shrinkage 
being  oniy  in  their  thick' 
ness ;    neither    will    tim 
ber  shrink  on  end  or  in 
Fig.  132.  jj.g  length.     Figures   132 

and  133  show  how  timber  will  shrink.  The 
first  from  a  quartered  log,  the  other  from  one  with  parallel  cuts. 
The  dotted  part  shows  the  shrinkage.  The  side-pieces  G  in  Fig.  133 
will  curl,  as  shown,  besides  shrinking.  By  observing  the  directions 
of  the  annular  rings,  therefore,  the  future  behavior  of  the  timber  can 
be  readily  predicted.  Of  course,  the  figures  are  greatly  exaggerated 
to  show  the  effect  more  clearly. 

If  the  heart  is  not  straight  its  entire  length,  the 
piece  will  twist  lengthwise.  Shrinkage  is  a  serious 
danger,  but  the  chief  danger  in  the  use  of  timber  lies  in  its  decay. 
All  timber  will  decay  in  time,  but  if  it  is  properly  dried,  before  be- 
ing built  in,  and  all  sap-wood  discarded,  and  then  so  placed  that  no 
moisture  can  get  to  the  timber,  while  fresh  air  has  access  to  all  parts 
of  it,  it  will  last  for  a  very  long  time ;  some  woods  even  for  many 
centuries.  In  proportion  as  we  neglect  the  above  rules,  will  its  life 
be  short-lived.  There  are  two  kinds  of  decay,  wet  and  dry  rot.  The 
wet  rot  is  caused  by  alternating  exposures  to  dampness  and  dryness ; 
or  by  exposure  to  moisture  and  heat ;  the  dry-rot,  by  confining  the 
timber  in  an  air-tight  place.  In  wet  rot  there  is  "  an  excess  of  evap- 
oration ;  "  in  dry  rot  there  is  an  "  imperfect  evaporation."  Beams 
with  ends  built  solidly  into  walls  arc  apt  to  rot ;  also  beams  sur- 
rounded solidly  with  fire-proof  materials  ;  beams  in  damp,  close,  and 
imperfectly  ventilated  cellars ;  sleepers  bedded  solidly  in  damp  mor- 
tar or  concrete,  and  covered  with  impervious  papers  or  other  mate- 
rials; also  timbers  exposed  only  at  intervals  to  water  or  dampness,  or 
timbers  in  "  solid  "  timbered  floors. 

Dry  rot  is  like  a  contagious  disease,  and  will  gradually  not  only 
eat  up  the  entire  timber,  but  will  attack  all  adjoining  sound  wood- 
work.    Where  rotted  woodwork  is  removed,  all  adjoining  woodwork. 


DECAY    OF    TIMBER.  191 

masonry,  etc.,  should  be  thoroughly  scraped  and  washed  with  strong 
acids. 

Ventilation  Vvherc  wood  lias,  of  necessity,  to  be  surrounded 

necessary,  ^^.j^j  fireproof  materials,  a  system  of  pipes  or  other 
arrangements,  should  be  made  to  force  air  to  same  through  holes, 
either  in  the  floors  or  ceilings,  but  in  no  case  connecting  two  floors  ; 
the  holes  can  then  be  made  small  enough  not  to  allow  the  passage  of 
fire.  Where  the  air  is  forced  in  under  pressure  it  would  be  advisa- 
ble at  times  to  force  in  disinfectants,  such  as  steam  containing  evap- 
orated carbolic  acid,  fumes  of  sulphur,  etc. 

Coating  woodwork  Avith  paint  or  other  preparations  will  only  rot 
the  Avood,  unless  it  has  been  first  thoroughly  dried  and  every  particle 
of  sap  removed. 

Cross-bridging.  Timber  must  not  be  used  too  thin,  or  it  will  be  apt 
to  twist.  For  this  reason  floor-beams  should  not  be  used  thinner 
than  three  inches.  To  avoid  twisting  and  curling,  cross-bridging  is 
resorted  to.  That  is,  strips  usually  2"  X  3"  are  cut  between  the 
beams,  from  the  bottom  of  one  to  the  top  of  the  next  one,  the  ends 
being  cut  (in  a  mitre-box),  so  as  to  fit  accurately  against  the  sides  of 
beams,  and  each  end  nailed  with  at  least  two  strong  nails.  The  strips 
are  always  placed  in  double  courses,  across  the  beams,  the  courses 
crossing  each  other  like  the  letter  x  between  each  pair  of  beams. 

This  is  known  as  "herring-bone"  cross-bridging.  Care  should 
be  taken  that  all  the  parallel  pieces  in  each  course  are  in  the  same 
line  or  plane.  The  lines  of  cross-bridging  can  be  placed  as  frequently 
as  desired,  for  the  more  there  are,  the  stilfer  will  be  the  floor.  About 
six  feet  between  the  lines  is  a  good  average.  Sometimes  sohd  blocks 
are  used  between  the  beams,  in  place  of  the  herring-bone  bridging. 
Cross-bridging  is  also  of  great  help  to  a  floor  by  relieving  an  individ- 
ual beam  from  any  great  weight  accidentally  placed  on  it  (such  as 
one  leg  of  a  safe,  or  one  end  of  a  book-case),  and  distributing  the 
weight  to  the  adjoining  beams.  Unequal  settlements  of  the  individ- 
ual beams  are  thus  avoided.  Where  a  floor  shows  signs  of  weakness, 
or  lacks  stiffness,  or  where  it  is  desirable  to  force  old  beams,  that 
St  ffenin''  cannot  be  Avell  removed,  to  do  more  work,  two  lines 

weak  floors,  of  slightly  wedge-shaped  blocks  are  driven  tightly 
between  the  beams,  in  place  of  the  cross-bridging.  The  beams  are 
then  bored,  and  an  iron  rod  is  run  between  the  lines  of  wedges,  from 
the  outer  beam  at  one  end  to  the  outer  beam  at  the  other,  and,  of 
course,  at  right  angles  to  all.     At  one  end  the  rod  has  a  thread  nnd 


192  SAFK    liUlLDIxa. 

nut,  and  by  screwing  up  the  latter  the  beams  are  all  forced  upwards, 
"  cambered,"  and  the  entire  floor  arched.  It  will  be  found  much 
stronger  and  stiifer ;  but,  of  course,  will  need  levelling  for  both  floor 
and  ceiling.  Under  the  head  and  nut  at  ends  of  rod,  there  must  be 
ample  washers,  or  the  sides  of  end  beams  will  be  crushed  in,  and  the 
effect  of  the  rod  destroyed. 

Girders,  which  cannot  be  stiffened  sideways,  should  be,   at  least, 
half  as  thick  as  they  are  deep,  to  avoid  lateral  flexure. 

•  In  using  wooden  beams  and  girders,  much  fram- 

beams.  ing  has  to  be  resorted  to.  The  used  joints  between 
timbers  are  numerous,  but  only  a  very  few  need  special  mention 
here.  Beams  should  not  rest  on  girders,  if  it  can  be  avoided,  on  ac- 
count of  the  additional  dropping  caused  by  the  sum  of  the  shrinkage 
of  both,  where  one  is  over  the  ■  other.  If  framing  is  too  expensive, 
bolt  a  wide  piece  to  the  under  side  of  the  girder,  sufliciently  wider 
than  the  p-irder  to  allow  the  beams  to  rest  on  it,  each  side.  If  this 
is  not  practicable  bolt  pieces  onto  each  side  of  the  girder,  at  the  bot- 
tom, and  notch  out  the  beams  to  rest  against  and  over  these  pieces. 
The  bearin"-  of  a  beam  should  always  be  as  near  its  bottom  as  possible. 
If  a  beam  is  notched  so  as  to  bear  near  its  centre,  it  will  split  longi- 
tudinally. Where  a  notch  of  more  than  one-third  the  height  of 
beam,  from  the  bottom,  is  necessary,  a  wrought-iron  strap  or  belt 
should  be  secured  around  the  end  of  beam,  to  keep  it  from  splitting 
lengthwise. 

If  framing  can  be  used,  the  best  method  is  the  "  tusk  and  tenon  " 
joint,  as  shoAvn  in  Figs.  134  and  135.     In  the  onecase  the  tenon  goes 
through  the  girder  and  is  secured  by  a  wooden  wedge  on  the  other 
side;  in  the  other   it  goes 
in  only  about  a  length  equal 
to   twice  its  depth,  and   is 
spiked  from  the  top  of  gir- 
der.    The  latter  is  the  most 
^'^' '^'*"  used.     By  both  methods  the 

girder  is  weakened  but  very  little,  the  principal 
cut  being  near  its  neutral  axis,  while  the  beam  gets  bearing  near  its 
bottom,  and  its  tenon  is  thoroughly  strengthened  to  prevent  its  shear- 
ing off.  The  dimensions  given  in  the  figures  are  all  in  parts  of  the 
height  of  beams.  Headers  and  trimmers  at  fire-places  and  other 
openings  are  frequently  framed  together,  though  it  would  be  more 
advisable  to  use  "  stirrup-irons."  The  short  tail-beams,  however,  can 
be  safely  tenoned  into  the  header. 


STIERUP-inOXS.  103 

In  calculating  the  strength  of  framed  timber,  the  point  where  the 
mortise,  etc.,  are  cut,  should  be  carefully  calculated  by  itself,  as  the 
cutting  frequently  renders  it  dangerously  weak,  at  this  point,  if  not 
allowed  for.  For  the  same  reason  plumbers  should  not  be  allowed 
to  cut  timbers.  As  a  rule,  however,  cuts  near  the  wall  are  not  dan- 
gerous, as  the  beam  being  of  uniform  size  throughout,  there  is  usu- 
ally an  excess  of  strength  near  the  wall. 

Stirrup-irons<  Stirrup-irons  are  made  of  wrought-iron  ;  they  are 

secured  to  one  timber  in  order  to  provide  a  resting-place  for  another 
timber,  usually  at  right  angles  to  and  carried  by  the  former.  They 
should  always  lap  over  the  farther  side 
of  the  carrying  timber,  to  prevent  slip- 
ping, as  shown  in  Fig.  13G. 

The  iron  should  be  sufficiently  wide 

not  to  crush  the  beam,  where  resting  on 

it';  the  section  of  iron  must  be  sufficient 

not  to  shear  off   each   side   of    beams. 

The  twist  must  not  be  too   sudden,  or 

it  will  straighten  out  and  let  the  carried  '^'    "  * 

timber  down.     To  put  the  above  in  formulae  we  should  have  : 

for  the  width  of  stirrup-iron  {x) 

s 

Width  of  ^=  J    /  c\  r69) 

Stirrup-irons.  "   '  —  '-  -     ' 


(y)' 


/         _ 
Where  x  =  the  width  of  stirrup-iron,  in  inches. 

"Where  s  =  the  shearing  strain,  in  lbs.,  on  end  of  beam,  being  car- 
ried. 

Where  b  ==  the  width  of  beam  being  carried,  in  inches. 

Where  (-TT  )  =  the   safe  resistance,  in   pounds,  to  compression, 

across  the  fibres,  of  the  beam,  being  carried. 

For  the  thickness  of  stirrup-iron  we  should  have : 
s 


y- 


2.x.{f) 


(7D) 


Which  for  wrought-iron  (Table  IV.)  becomes, 


Thickness  of  y= f (71) 

Stirrup-iron.     "^        IGOOO.  a; 

Where  7/z=rthe  thickness  of  stirrup-iron,  in  inches. 
Where  s  =  the  shearing  strain  on  end  of  beam,  in  lbs. 
Where  a:=:is  found  by  formula  (G9). 


104  SAKE    15UILDIXG. 

Providing,  however,  that  y  should  ncvci-  be  less  than  one-f|uarter 
inch  thick. 

Example. 

A  girder  carries  the  end  of  a  beam,  on  tohich  there  is  a  uniform  load 
of  two  thousand  pounds.  The  beam  is  four  inches  thick,  and  of 
Georgia  pine.  What  size  must  the  stirrup-iron  be? 
bxample  stir-  The  shearing  strain  at  each  end  of  the  beam  will, 
rup-irors.  of  course,  be  one  thousand  pounds,  which  will  be  the 
load  on  stirrup-irons.  (See  Table  VII).  From  Table  IV  we  find 
for  Georgia  pine,  across  the  fibres,  (-^)  =200,  we  have,  therefore, 

for  the  width  of  stirrup-iron  from  Formula  (69) 
1000  _, 
^  —  4.200       ^ 
Therefore  the  thickness  of  iron  from  Formula  (71)  should  be 
_      1000      _    1    „ 
^"iGOOO.li         20 
we  must  make  the  iron  however  at  least  \'  thick  and  therefore  use  a 
section  of  Ij  X  i"- 

In  calculating  ordinary  floor-beams  the  shearing  strain  can  be 
overlooked,  as  a  rule ;  for,  in  calculating  transverse  strength  we 
allow  only  the  safe  stress  on  the  fibres  of  the  upper  and  lower  edges, 
while  the  intermediate  fibres  are  less  and  less  strained,  those  at  the 
neutral  axis  not  at  all.  The  reserve  strength  of  these  only  partially 
used  fibres  Avill  generally  be  found  quite  ample  to  take  up  the  shear- 
ing strain. 

Rectangular  The   formulae    for  transverse   strength  are   quite 

beams,  complicated,  but  for  rectangular  sections  (wooden 
beams)  they  can  be  very  much  simplified  provided  we  are  calcu- 
lating for  strength  only  and  7iot  taking  deflection  into  account. 

Keraembering  that  the  moment  of  resistance  of  a  rectangular  sec- 

b  d'^ 
tion  is  (Table  I)  =~7r-  and  inserting  into  Formula  (18)  the  value 

for  m  according  to  the  manner  of  loading  and  taken  from  (Table 
VII),  we  should  have  : 

For  uniform  load  on  beam. 
Transverse  i,  tj       /  7.  \ 

strength  of  u  — -^  .  ( -~\  (72) 

rectangu-  9.L       \// 

lar  beams. 


For  centre  load  on  beam. 


lS-(7)  '''' 


KECTANGUl.AK    IJEA.MSJ.  195 

For  load  at  any  point  ufbeam. 

For  unl/ot  m  load  on  cantilever. 

For  load  concentrated  at  end  of  cantilecer. 


r2.L  V  / 
For  load  at  any  point  of  cantilever. 

b.d^  _ 

72.Y  •  V  / 


^,-^;l^,.(J^)  (77) 


"Where  u  =  safe  uniform  load,  in  pounds. 

Where  w=  safe  centre  load  on  beam,  in  pounds;  or  safe  load  at 
end  of  cantilever,  in  pounds. 

Where  ?«,=  safe  concentrated  load,  in  pounds,  at  any  point. 

Where  V=  length,  in  feet,  from  wall  to  concentrated  load  (in  can- 
tilever). 

Where  M  and  iV=  the  respective  lengths,  in  feet,  from  concen- 
trated load  on  beam  to  each  support. 

Where  L=  the  length,  in  feet,  of  span  of  beam,  or  length  of  canti- 
lever. 

Where  b  =  the  breadth  of  beam,  in  inches. 

Where  d  =  the  depth  of  beam,  in  inches. 

Where('— )=the  safe   modulus  of   rupture,  per  square  inch,  of 

the  material  of  beam  or  cantilever  (see  Table  IV). 

The  above  formulse  are  for  rectangular  wooden  beams  supported 
against  lateral  flexure  (or  yielding  sideways).  Where  beams  or  gir- 
ders are  not  supported  sideways  the  thickness  should  be  equal  to  at 
least  half  of  the  depth. 

No  allowance  ^''^^  «^»^'«  formidce  make  no  allowance  for  dejlec- 
for  deflection,  tion,  and  except  in  cases,  such  as  factories,  etc., 
where  strength  only  need  be  considered  and  not  the  danger  of  crack- 
ing plastering,  or  getting  floors  too  uneven  for  machinery,  are  really 
of°but  little  value.  They  are  so  easily  understood  that  the  simplest 
example  will  answer : 

Example. 
Take  a  3"  X  10"  hemlock   timber   and  9  feet  long   {clear  span), 
loaded  in  different  ways,  what  will  it  safely  carry  ?  taking  no  account 
of  deflection. 


190  SAFE    BUILDING. 

The  safe  modulus  of  rupture  (  A  )  for  hemlock  from  Table  IV  is 

=  750  pounds. 
If  both  ends  are  supported  and  the  load  is  uniformh'  distributed 
the  beam  will  safelj'  carry,  (Formula  72)  : 

u=  ^^-^-  750  =:r  2778  pounds. 

If  both  ends  are  supported  and  the  load  concentrated  at  the  centre, 
the  beam  will  safely  carry,  (Formula  73)  : 

w  =  -^^  .  750  =  1 389  pounds. 
18.9  '■ 

If  both  ends  are  supported  and  the  load  is  concentrated  at  a  point 

I,  distant  four  feet  from  one  support  (and  five  feet  from  the  other) 

the  beam  will  safely  carry,  (Formula  74)  : 

■      ■       750  =  1406  pounds. 


72.4.5 

If  one  end  of  the  timber  is  built  in  and  the  other  end  free  and  the 

load  uniformly  distributed,  the  cantilever  will  safely  carry,  (Formula 

75):  '  , 

"10- 
u  =  — —  .  750  =  G94  pounds. 

If  one  end  is  built  in  and  the  other  end  free,  and  the  load  concen- 
trated at  the  free  end,  the  cantilever  will  safely  carry,  (Formula  7G)  : 

3.10-    .-„       o,-  1 

w=- — -- .  (iiO^o-iii  pounds. 

If  one  end  is  built  in  and  the  other  end  free,  and  the  load  concen- 
trated at  a  point  I,  which  is  5  feet  from  the  built-in  end,  the  canti- 
lever will  safely  carry,  (Formula  77)  : 

i«,  =  ^-^-  .  750  =  G25  pounds. 

Where,  however,  the  span  of  the  beam,  in  feet,  greatly  exceeds 
the  depth  in  inches  (see  Table  VIII),  and  regard  must  be  had  to  de- 
flection, the  formulae  (28)  and  (29),  also  (37)  to  (42)  should  always 
be  used,  inserting  for  i  its  value  from  Table  I,  section  Xo.  2,  or :  — 

Where  b  =  the  thickness  of  timber,  in  inches. 

Where  d  =  the  depth  of  timber,  in  inches. 

Where  i  =  the  moment  of  inertia  of  the  cross-section,  in  inches. 

Table  IX,  however,  gives  a  much  easier  method  of  calculating 
wooden  beams,  allowing  for  both  rupture  and  deflection,  and  For- 
muliB  (72)  to  (77)  have  only  been  given  here,  as  they  are  often  erro- 


vhE   OF   TAUI.KS.  197 

neously  given  in  text-books,  as  tlie  only  calculations  necessary  for 
beams. 

Basis  of  Tables      '^"^  ^^'^^^  furtlier  simplify  to  the  architect  the  labor 
XII  and  XIII.    of  calculating  wooden  beams   or  girders,  the  writer 
has  constructed  Tables  XII  and  XIII. 

Table  XII  is  calculated  for  floor  beams  of  dwellings,  offices' 
churches,  etc.,  at  90  pounds  per  square  foot,  including  weight  of 
construction.     The  beams  are  supposed  to  be  cross-bridged. 

Table  XIII  is  for  isolated  girders,  or  lintels,  uniformly  loaded, 
and  supported  sideways.  When  not  supported  sideways  decrease  the 
load,  or  else  use  timber  at  least  half  as  thick  as  it  is  deep.  In  no  case 
will  beams  or  girders  (with  the  loads  given)  deflect  sudiciently  to 
crack  plastering.  For  convenience  Table  XII  has  been  divided  into 
two  parts,  the  first  part  giving  beams  of  from  5'  0"  to  15'  0"  sjian, 
the  second  part  of  from  15'  0"  to  29'  0"  span. 

How  to  use  '^^'*-*  "*®  ^^  ^^^^  table  is  very  simple  and  enables  us 

Table  XII.  to  select  the  most  economical  beam  in  each  case. 
Foi-  instance,  we  have  say  a  span  of  21'  G".  We  use  the  second  part 
of  Table  XII.  The  vertical  dotted  line  between  21'  0"  and  22'  0"  is, 
of  course,  our  line  for  21'  6".  We  pass  our  finger  down  this  line  till 
we  strike  the  curve.  To  the  left  opposite  the  point  at  which  we 
struck  the  curve,  we  read : 

21. G  spruce,  W.  P.  5G  —4-14-14  or: 
at  21'  G'  span  we  can  use  spruce  or  white-pine  floor  beams,  of  56 
inches  sectional  area  each,  viz. :  4"  thick,  14"  deep  and  14''  from  cen- 
tres. Of  coiu'se  we  can  use  any  other  beam  heloiu  this  point,  as  they 
are  all  stronger  and  stiffer,  but  we  must  not  use  any  other  beam 
above  this  point.  Now  then,  is  a  4"  X  li"  beam  of  spruce  or  white 
pine,  and  14"  from  centres  the  most  economical  beam.  TV'e  pass  to 
the  columns  at  the  right  of  the  curve  and  there  read  in  the  first 
column  48".  This  means  that  while  the  sectional  area  of  the  beam 
is  56  square  inches,  it  is  equal  to  only  48  S(|uare  inches  per  square 
foot  of  floor,  as  the  beams  are  more  than  one  foot  fi'om  centres.  In 
this  column  the  areas  are  all  reduced  to  the  '•  area  per  scjuare  foot  of 
floor,"  so  that  we  can  see  at  a  glance  if  there  is  any  cheaper  beam 
helow  our  point.  We  find  below  it,  in  fact,  many  cheaper  beams,  the 
smallest  area  (per  square  foot  of  floor)  being,  of  course,  the  most 
economical.  The  smallest  area  we  find  is  3G,  0  or  3G  scjuare  inches 
of  section  per  square  foot  of  floor  (this  we  find  three  times,  in  the 
sixteenth,  twenty-ninth  and  thirty-first  lines  from  the  bottom).  Pass- 
ing to  the  left  we  find  they  represent,  respectively,  a  Georgia  pine 


198  SAFE    BUILDING. 

beam,  3"  thick,  IG"  deep  and  IG"  from  centres;  or  a  Georgia  pine 
beam  3"  thick,  14"  deep  and  14"  from  centres;  or  a  -white  oak  beam 
S"  thick,  16"  deep  and  16"  from  centres.  If  therefore,  we  do  not 
consider  depth,  or  distance  from  centres,  it  would  simply  be  a  ques- 
tion, which  is  cheaper,  48  inches  (or  four  feet  "board-measure")  of 
white  pine  or  spruce,  or  36  inches  (or  three  feet  "board-measure") 
of  either  white  oak  or  Georgia  pine.  The  four  other  columns  on 
the  right-hand  side,  are  for  the  same  purpose,  only  the  figures  for 
each  kind  of  wood  are -in  a  column  by  themselves  ;  so  that,  if  we  are 
limited  to  any  kind  of  wood  we  can  examine  the  figures  for  that 
wood  by  themselves.  Take  our  last  case  and  suppose  we  are 
limited  to  the  use  of  hemlock;  now  from  the  point  where  our  verti- 
cal line  (21'  G")  first  struck  the  curve,  we  pass  to  the  right-hand  side 
of  Table,  to  the  second  column,  which  is  headed  "  Hemlock."  From 
this  point  we  seek  the  smallest  figure  below  this  level,  but  in  the  same 
column;  we  find,  that  the  first  figure  we  strike,  viz:  41,  2  is  the 
smallest,  so  we  use  this ;  passing  along  its  level  to  the  left  we  find  it 
represents  a  hemlock  beam  of  48  square  inches  cross-section,  or  3" 
thick,  IG"  deep  and  14"  from  centres. 

In  case  the  size  of  the  beam  is  known,  its  safe  span  can,  of  course, 
be  found  by  reversing  the  above  procedure,  or  if  the  depth  of  beam 
and  span  is  settLd,  we  can  find  the  necessary  thickness  and  distance 
between  centres;  in  this  way  the  Table,  of  course,  covers  every 
problem. 

Table  XIII  is  calculated  for  wooden  girders  of  all  sizes.  Any 
thickness  not  given  in  the  table  can  be  obtained  by  taking  the  line 
for  a  girder  of  same  depth,  but  one  inch  thick  and  multiplying  by  the 
thickness.  For  very  short  spans,  look  out  for  danger  of  horizontal 
shearing  (see  formula  13)  ;  where  this  danger  exists,  pass  vertical 
bolts  through  ends  of  girder,  or  bolt  thin  iron  plates,  or  straps,  or 
even  boards  with  vertical  grain,  to  each  side  of  girder,  at  ends. 
How  to  use  The  use  of  this  table  is  very  simple.     The  vertical 

Table  XIII.  columns  to  the  left  give  the  safe  uniform  loads  on 
girders  (sufficiently  stiff  not  to  crack  plastering)  for  different  woods: 
these  apply  to  the  dotted  parts  of  curves.  The  columns  on  the  right- 
hand  side  give  the  same,  but  apply  to  the  parts  of  curves  drawn  in 
full  lines. 

If  we  have  a  G"  X  16"  Georgia  pine  beam  of  20  feet  span  and 
want  to  know  what  it  will  carry,  Ave  select  the  curve  marked  at  its 
upper  end  6  X  16  =  96;  we  follow  this  curve  till  it  intersects  the 
vertical  line  20'  0";  as  this  is  in  the  part  of  curve  drawn  full,  we 


TAIil.K    XV.  U»0 

ir>()X   I-BEAM    GIUDKKS,  BUACEI)    SIDEWAYS. 


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For  Ste^l  Bpams,  add  one-quarter  to  safe  uniform  load  on  Iron  Beams  ;  but  length 
of  span  in  feet  inu<t  not  exceed  twice  the  depth  of  beam  in  inches,  or  defleclioii  will 
be  too  great  for  plastering. 

IKOX    BEAM    TABLES.  201 

pass  horizontally  to  the  right  and  find  under  the  column  marked 
"  Georgia  Pine,"  7980,  which  is  the  safe,  uniform  load  in  pounds.  Sup- 
posing, however,  we  had  simjjly  settled  the  span,  say  8  feet,  and  load, 
say  7000  pounds,  and  wished  to  select  the  most  economical  girder, 
being,  we  will  say,  limited  to  the  use  of  white  pine :  the  span  not 
being  great  we  will  expect  to  strike  the  dotted  part  of  curve,  and 
therefore  select  the  fourth  (white  pine)  column  to  the  left.  We  pass 
down  to  the  nearest  figure  to  7000  and  then  pass  horizontally  to  the 
right  till  we  meet  the  vertical  8  feet  line ;  this  we  find  is,  as  we  ex- 
pected, at  the  dotted  part,  and  therefore  our  selection  of  the  left 
^  column  was  right.  "We  follow  the  curve  to  its  upper  end  and  find  it 
requires  a  girder  4"  X  12"  =  48  square  inches.  Now,  can  wo  use  a 
cheaper  girder  ?  of  course,  all  the  lines  under  and  to  the  right  of  our 
curve  are  stronger,  so  that  if  either  has  a  smaller  sectional  area,  we 
will  use  it.  The  next  curve  we  find  is  a  6"  X  10"  =  GO";  then 
comes  a  4"  X  14"  =  56";  then  an  8"  X  10"  =  80";  then  a  G"  X 
12"=  72"  and  so  on;  as  none  has  a  smaller  area  we  will  stick  to  our 
4"  X  1 2"  girder,  provided  it  is  braced  or  supported  sideways.  If  not, 
to  avoid  twisting  or  lateral  flexure,  we  must  select  the  next  cheapest 
section,  where  the  thickness  is  at  least  equal  to  half  the  depth;  i  the 
cheapest  section  beyond  our  curve  that  corresponds  to  this,  we  find 
is  the  6"  X  10"  girder,  which  we  should  use  if  not  braced  sideways. 
In  the  smaller  sections  of  girders  where  the  difference  between 
the  loads  given  from  line  to  line  is  proportionally  great,  a  safe  load 
should  be  assumed  between  the  two,  according  to  the  proximity  to 
either  line  at  which  the  curve  cuts  the  vertical.  The  point  where 
the  curve  cuts  the  lotlom  horizontal  line  of  each  part  is  the  lenf^th  of 
span  for  which  the  safe  load  opposite  the  line  is  calculated. 
Heavier  Floors.  Where  a  different  load  than  90  pounds  per  square 
foot,  must  be  provided  for,  we  can  either  increase  the  thickness  of 
beams  as  found  in  Table  XIT,  or  decrease  their  distance  from  cen- 
tres, either  in  proportion  to  the  additional  amount  of  load.  Or,  if 
we  wish  to  be  more  economical,  we  can  calculate  the  safe  uniform 
load  on  each  floor  beam,  and  consider  it  as  a  separate  girder,  sup- 
ported sideways,  using  of  course.  Table  XIII. 

Basis  of  Tables       The  Tables  XIV  and  XV  are  very  similar  to  the 

XIV  and  XV.  foregoing,  but  calculated  for  wrought-iron   I-beams. 

Table  XIV  gives  the  size  of  beams  and  distance  from  centres  re- 

quired  to  carry  different  loads  per  square  foot  of  floor,  150  pounds 

'  The  rule  for  calculating  the  exact  thickness  will  be  found  later,  Formula  (78). 


202  SAFE    BUILDING. 

per  square  foot  of  floor  (including  the  weight  of  construction),  ho-w- 
ever,  being  the  usual  load  allowed  for  in  churches,  office-buildings, 
public  halls,  etc.,  where  the  space  between  beams  is  filled  with  arched 
brickwork,  or  straight  hollow-brick  arches,  and  then  covered  over 
with  concrete.  A  careful  estimate,  however,  should  be  made  of 
the  exact  weight  of  construction  per  square  foot,  including  the  iron- 
work, and  to  this  should  be  added  70  pounds  per  square  foot,  which 
is  the  greatest  load  likely  ever  to  be  produced  if  packed  solidly  with 
people.  Furniture  rarely  weighs  as  much,  though  heavy  safes  should 
be  provided  for  separately.  The  load  on  roofs  should  be  30  pounds 
additional  to  the  weight  of  construction,  to  provide  for  the  weight  of 
snow  or  wind.  Look  out  for  tanks,  etc.,  on  roofs.  Plastered  ceil- 
inss  hanging  from  roofs  add  about  10  pounds  per  square  foot,  and 
slate  about  the  same.  "Where  a  different  load  than  given  in  the 
Table  must  be  provided  for,  the  distance  between  centres  of  beams 
can  be  reduced,  proportionally  from  the  next  greater  load ;  or  the 
weight  on  each  beam  can  be  figured  and  the  beam  treated  as  a  girder, 
supported  sideways,  in  that  case  using  Table  XV.  Both  tables  are 
calculated  for  the  beams  not  to  deflect  sufficiently  to  crack  plastering. 
How  to  use  The  use  of  Table  XIV  is  very  simple.     Supposing 

Table  XIV.  ^ve  have  a  span  of  24  feet  and  a  load  of  150  pounds 
per  square  foot.  We  pass  down  the  vertical  line  24'  0"  and  strike 
first  the  12" -96  pounds  beam,  which  (for  150  pounds)  is  opposite 
(and  three-quarter  way  between)  3'  0"  and  3'  4"  therefore  3'  3"  from 
centres.  The  next  beam  is  the  12"- 120  pounds  beam  4'  0"  from 
centres;  then  the  12"-125  pounds  beam  4'  1"  from  centres;  then 
the  15"-  125  pounds  beam  5'  0"  from  centres  and  so  on.  It  is  sim- 
ply a  question,  therefore,  which  "  distance  from  centres  "  is  most  de- 
sirable and  as  a  rule  in  fireproof  buildings  it  is  desirable  to  keep 
these  as  near  alike  as  possible,  so  as  not  to  have  too  many  different 
spans  of  beam  arches  and  centres.  If  economy  is  the  only  question, 
we  di\  ide  the  weight  of  beam  by  its  distance  from  centres,  and  the 
curve  giving  the  smallest  result  is,  of  course,  the  cheapest.  Sup- 
posing, however,  that  we  desire  all  distances  from  centres  alike,  say 
5  feet.  In  that  case  we  pass  down  the  150-pound  column  to  and 
then  along  the  horizontal  line  5'  0"  till  we  strike  the  vertical 
(span)  line,  in  this  case  24'  0",  and  then  take  the  cheapest 
beam  to  the  right  of  the  point  of  intersection.  Thus,  in  our 
case  the  nearest  beam  would  be  15"- 125  pounds;  next  comes  12^" 
-170  pounds;  then  15" -150  pounds,  etc.  As  the  nearest  beam  is 
the  lightest  in  this  case,  we  should  select  it.     The  weight  of  a  beam 


USE    OF    STEEL.  203 

is  always  given  per  yard  of  Icngtli.  The  reason  for  this  is  that  a 
square  inch  of  wrought-iron,  one  yard  long,  weighs  exactly  10 
pounds.  Therefore  if  we  know  the  Aveight  per  yard  in  pounds  we 
divide  it  by  ten  to  obtain  the  exact  area  of  cross-section  in  square 
inches ;  or  if  we  know  the  area,  we  multiply  by  ten  and  obtain  the 
exact  weight  per  yard. 

Howtouss  '^^^^  "s^  of  Table  XV,  is  very  similar  to  that  of 

Table  XV.  Tabic  XlII,  but  that  the  safe  uniform  load  is  given 
(in  the  first  column)  in  tons  of  2000  pounds  each.  The  continuation 
of  the  two  20"  beams  up  to  42  feet  span  is  given  in  the  separate 
table,  in  the  lower  right-hand  corner.  To  illustrate  the  Table :  if  we 
have  a  span  of  say  21  feet  we  pass  down  its  vertical  line ;  the  first 
curve  we  strike  is  the  10^" -90  pounds  beam,  which  is  three-quarter 
space  beyond  the  horizontal  line  5  (tons);  therefore  a  10^"- 90 
pounds  beam  at  21  feet  span  will  carry  safely  5|  tons  uniform  load, 
and  will  not  deflect  sufficiently  to  crack  plaster.  (Each  full  horizontal 
space  represents  one  ton).  The  next  beam  at  21  feet  span  is 
10|"-105  pounds,  which  will  safely  carry  G^  tons.  Then  comes 
the  12" -96  pounds  beam,  which  will  safely  carry  7  tons,  and  so 
on  down  to  the  20" -272  pounds  beam,  which  will  safely  carry  33| 
tons. 

If  we  know  the  span  (say  1 7  feet)  and  uniform  load  (say  7^  tons) 
to  be  carried,  we  pass  down  the  span  line  1 T  0"  and  then  horizon- 
tally along  the  load  line  7^  till  they  meet,  which  in  our  case  is  at  the 
9" -125  pounds  beam;  we  can  use  this  beam  or  any  cheaper  beam, 
whose  curve  is  under  it.  We  pass  over  the  different  curves  under 
it,  and  find  the  cheapest  to  be  the  12"-9G  pounds  beam,  which  we, 
of  course,  use. 

Iron  beams  must  be  scraped  clean  of  rust  and  be  well  painted. 
They  should  not  be  exposed  to  dampness,  nor  to  salt  air,  or  they  will 
deteriorate  and  lose  strength  rapidly. 

Steel  beams.  Steel  beams  are   coming   into  use   quite   largely. 

They  are  cheaper  to  manufacture  than  iron  beams,  as  they  are  made 
directly  from  the  pig  and  practically  in  one  process ;  while  with  iron 
beams  the  ore  is  first  convi?rted  into  cast  iron,  then  puddled  into  the 
muck-bar,  re-heated,  and  then  rolled.  Steel  beams,  however,  are  not 
apt  to  be  of  uniform  quality.  Some  may  be  even  very  brittle  ;  they  are, 
however,  very  much  stronger  than  iron  (fully  25  per  cent  stronger),  but 
as  their  deflection  is  only  about  7,  3  per  cent  less  than  that  of  iron 
beams,  there  is  but  very  little  economy  of  material  possible  in  their  use. 
If  steel  beams  are  used  they  can  be  spaced  cue  (juarter  distance  (be- 


204  SAFE   BUILDING. 

twcen  centres)  farther  apart  than  given  in  Table  XIV  for  iron  beams ; 
or  they  will  safely  carry  one  quarter  more  load  than  given  in  Table 
XV ;  but  in  no  case,  where  full  load  is  allowed,  must  the  span  in  feet, 
(of  steel  beams),  exceed  twice  the  depih  in  inches.  With  full  safe 
loads  the  deflection  of  steel  beams  will  always  be  greater  than  that 
of  iron  beams  (about  ^  larger).  Where,  therefore,  it  is  desirable  not 
to  have  a  greater  deflection  than  with  iron  beams,  add  only  7^  per 
cent  to  the  distances  between  centres  or  "  safe  loads  "  as  given  in 
Tables  for  iron  beams,  instead  of  25  per  cent. 

Steel  beams  will  undoubtedly  supersede  iron  beams  before  many 
years  have  passed;  but  in  the  present  state  of  their  manufacture 
their  use  is  hardly  to  be  recommended.  Their  strength  and  con- 
sistency is  very  variable.  It  has  been  found  in  some  cases  that  steel 
beams  broke  suddenly  when  jarred,  (that  is,  were  very  brittle,) 
though  test  pieces  o£E  the  ends  of  these  same  beams  gave  very  satis- 
factory results.  If  steel  is  used,  not  only  should  samples  of  each 
piece  be  carefully  tested,  for  tenacity,  ductility,  elasticity,  elongation, 
etc.,  but  the  whole  beam  itself  should  be  tested  by  actual  loading.  It 
will  be  readily  seen  that  the  expense  of  such  tests  would  ])Rr  the  use 
of  steel,  but  no  architect  can  afford  to  take  any  chances  in  such  an 
important  part  of  his  building. 

Many  writers  even  claim,  that,  "  within  the  elastic  limit,"  the  addi- 
tional stiffness  of  steel  over  iron  does  not  appear ;  and  that  it  is  only 
beyond  this  limit  that  steel  is  somewhat  stiffer  than  iron. 
Lateral  Flexure  ^^  using  iron  and  steel  beams  it  is  very  important 
in  beams,  that  they  be  supported  sidewa^'S,  so  as  not  to  yield 
to  lateral  flexure.  Where  the  beams  are  isolated  and  unsupported 
sideways,  the  safe  load  must  be  diminished.  Just  how  much  to  di- 
minish this  load  is  the  question.  The  practice  amongst  iron  workers 
is  to  consider  the  top  flange  as  a  column  of  the  full  length  of  the 
span,  obliged  to  yield  sideways,  and  with  a  load  equal  to  the  greatest 
strain  on  the  flange.  Modifying,  therefore,  Formula  (3)  to  meet  this 
view,  we  should  have  : 

Beams  not  __         w  (78) 

braced  •—  „   ^^ 

sideways.  l-+-~-r, — 

b- 

Where  w,  =  the  safe  load,  in  pounds,  on  a  beam,  girder,  lintel  or 
straight  arch,  etc,  unsupported  sideways. 

Where  to  ^=  the  safe  load,  in  pounds,  on  a  beam,  lintel  or  straight 
arch  supported  sideways. 


LATi:i:.\I.     FLKXL'UE. 


205 


Where  L  =  the  length  of  clear  sjjan,  in  feet,  that  beam,  etc.,  is 
unsupported  sideways. 

Where  b  =  the  least  breadth  in  inches  of  top  flange,  or  least 
thickness  of  beam,  lintel  or  arch. 

Where  y  =a,  constant,  as  found  in  Table  XVI. 

(In  place  of  to  we  can  use  r  =  the  moment  of  resistance  of  beam 
6ii])ported  sideways,  and  in  place  of  w,  we  use  r^  =  the  moment  of 
resistance  of  beam  7iot  supported  sideways.) 

The  above  practice,  however,  would  seem  to  diminish  the  weight 
unnecessarily,  particularly  where  the  beam,  girder,  etc.,  is  of  uniform 
section  throughout;  for  while  the  beam  in  that  case,  would,  be 
equally  strong  at  all  points,  it  would  be  strained  to  the  maximum 
compression  only  at  the  point  of  greatest  bending-moraent,  the  strain 
diminishing  towards   each  support,  where   the  compression   would 

TABLE   XYI. 

VALUE   OF    Y   IN    FORMULA    (78). 


Material  of  beam,  girder,  lintel, 
straight  arch,  etc. 


Cast  -  Iron 

Wrouglit-lron . 

Steel 

AVood 

Stone  

Brick 


Value  of  ?/  for 

girders,  beams, 

etc.,  of   i-aria- 

lie  cross-sec- 

tious. 


0,5184 
0,(1432 
0,034G 
0,5702 
3,4560 
5,7024 


Value  of  y  for  beams, 

girders,  lintels, 
straight  arches,  etc., 
of  uniform  cross-sec- 
tion tlironghout. 


0,.'304 
0,0192 
0,0154 
0,2534 
1,5300 
2,5344 


cease  entirely.  To  consider,  therefore,  the  whole  as  a  long  column 
carrying  a  weight  equal  to  this  maximum  compressive  strain,  seems 
unreasonable.  Box  has  shown,  however,  that  the  maximum  tendency 
to  deflect  laterally  is  Avhen  we  consider  the  top  flange  (or  top  half  in 
rectangular  beams,  lintels  and  straight  arches)  as  a  column  equal  to 
two  thirds  of  the  span  (unsupported  sideways)  loaded  with  a  weight 
ec^ual  to  one-third  of  the  greatest  compressive  strain  at  any  point. 
Tlys  greatest  compressive  strain  is  always  at  the  point  of  greatest 
bending  moment  (usually  the  centre  of  span),  and  is  equal  to  the 

area  of  top  flange,  multiplied  by  (  -f  )  ^-  In  case  of  plate  girders  the 
angle-irons  and  part  of  web  between  angle-irons  should  be  included 
in  the  area.     Box's  theory  is  given  in  Formula  (5)  ;  if  then  we  take 


1  This  is  not  quite  correct.   The  greatest  compressive  strain  is  really  a  little 
less,  as  will  be  explained  in  Vol.  II. 


206  SAii;  liUii.DiNG. 

one-third  of  this  "maximum  tendency  to  detlcct"  as  safe,  we  should 
have  the  same  Formula  as  (78)  but  with  a  smaller  value  for  y.     The 

Use  of  writer  would  recommend  using  the  larger  value  for 

Table  XVI.  y^  where,  as  in  plate  girders,  trusses,  etc.,  the  section 
of  top  flange  or  chord  is  diminished,  varying  according  to  the  com- 
pressive strain  at  each  point;  and  using  the  smaller  value  for  y, 
where  the  section  of  beam,  girder  or  top  chord  is  uniform  throughout. 

Thus  the  10|"-90  pounds  beam  at  20  feet  span  wili  safely  carry 
(if  supported  sideways)  a  uniform  load  of  5,9  tons  or  11800  pounds 
(see  Table  XV.)  The  width  of  flange  being  4^",  and  this  width  and 
its  thickness,  of  course,  being  uniform  throughout  the  entire  length 
of  beam,  we  use  the  smaller  value  for  y  (second  column)  and  have 
for  the  actual  safe  uniform  load,  if  the  beam  is  not  secured  against 
lateral  flexure : 

^  11800  _      11800 

'^''~1 -|-0,011J2.    20- ~  1 -|- 0,379 

11800       „.-_  ,  .  „„  ^ 

800/  pounds,  or  4,28  tons. 


1,379 
Had  we  used  the  larger  value  for  y  =  0,0432  we  should  have  had 

11800  11800       .„„,  1  o  iQ    . 

u\= — , = =:  6365    pounds,   or    3,18    tons, 

'       1  +  0,854       1,854  1  '  '  » 

which  closely  resembles  the  value  (3,29)  given  in  the  Iron  Com- 
panies' hand-books,  but  is  an  excessive  reduction  under  the  circum- 
stances. 

Doubled  Where  two  or  more  beams  are  used  to  carry  the 

Beams,  same  load,  as  girders  for  instance,  or  as  lintels  in  a 
wall,  they  should  be  firmly  bolted  together,  with  ca-t-iron  separa- 
tors between.  In  this  case  use  for  b  in  Formula  (78)  the  total 
width,  from  outside  to  outside  of  all  flanges,  and  including  in  b  the 
spaces  between.  The  separators  are  made  to  fit  exactly  between  the 
inner  sides  of  webs  and  top  and  bottom  flanges.  The  separator  is 
swelled  out  for  the  bolt  to  pass  through.  Sometimes  there  are  two 
bolts  to  each  separator,  but  it  is  better  (weakening  the  beam  less)  to 
have  but  one  at  the  centre  of  web.  The  size  of  separators  and  bolts 
vary,  of  course,  to  suit  the  different  sizes  of  beams.  They  should  be 
placed  apart  about  as  frequently  as  twenty  times  the  Avidth  of  flange 
of  a  single  beam.  Where  beams  are  placed  in  a  floor,  the  floor 
arches  usually  provide  the  side  bracing.  But  in  oi'der  to  avoid  un- 
Tie-rods.  equal  deflections,  and  possible  cracks  in  the  ai-ches, 
^from  unequal  or  moving  loads  or  from  vibrations)  and  also  to  take 
up  thfi  thru.5t  on  the  end  beams  of  each  floor,  it  is  necessary  to  place 


TllC    RODS.  207 

lines  of  tie-rods  across  tlie  entire  line  of  beams.  The  size  of  these 
rods  can  be  calculated  as  already  explained  in  the  Chapter  on 
Arches  (p.  166);  they  are  usually  made,  however,  from  f"  to  |" 
diameter.  P^ach  rod  extends  from  the  outside  web  of  one  beam 
to  the  outside  web  of  the  next  beam.  The  next  rod  is  a  little 
to  one  side  of  it,  so  that  the  rods  do  not  really  form  one  straight 
line,  but  every  other  rod  falls  in  the  same  line.  Care  must  be  taken 
not  to  get  the  rods  too  long,  or  there  will  have  to  be  several  washers 
under  the  head  and  nut,  making  a  very  unsightly  job,  to  say  the 
least.  Contractors  will  do  this,  however,  for  the  sake  of  the  con- 
venience of  ordering  the  rods  all  of  one  or  two  lengths.  Where, 
therefore,  the  beams  are  not  spaced  evenly  the  contractor  should  be 
warned  against  this.  One  end  of  the  rod  has  a  "  head  "  welded  on, 
the  other  has  a  "  screw-end,"  which  need  not  be  "  up-set ;"  the  nut  is 
screwed  along  this  end,  thus  forcing  both  nut  and  head  to  bear 
against  the  beams  solidly.  The  distance  between  lines  of  tie-rods, 
would  depend  somewhat  on  our  calculation,  if  made ;  the  usual  prac- 
tice, however,  is  to  place  them  apart  a  distance  equal  to  about 
twenty  times  the  width  of  flange  of  a  single  beam. 
Flitch-plate  Sometimes   where   wooden  girders  have  heavier 

Girder.  loads  to  carry  than  they  are  capable  of  doing,  and 
yet  iron  girders  cannot  be  afforded,  a  sheet  of  plate-iron  is  bolted  be- 
tween two  wooden  girders.  In  this  case  care  must  be  taken  to  so 
proportion  the  iron,  that  in  taking  its  share  of  the  load,  it  will 
deflect  ecpially  with  the  wooden  girders,  otherwise  the  bolts  would 
surely  shear  off,  or  else  cru=h  and  tear  the  wood. 

We  consider  the  two  wooden  girders  as  one  girder  and  calculate 
(or  read  from  Table  XIII)  their  safe  load,  taking  care  not  to  exceed 
0,03  inches  of  deflection  per  foot  of  span.  We  then,  from  Table  VII 
or  Formulse  (37)  to  (41)  obtain  the  exact  amount  of  their  deflection 
under  this  load.  We  now  calculate  the  iron  plate,  for  deflection 
only,  inserting  the  above  amount  of  deflection,  and  for  the  load  the 
balance  to  be  borne  by  the  iron-work.  An  example  will  best  illus- 
trate this : 

Example. 

A  Flitch-plate  girder  of  20-foot  span  consists  of  tico  Georgia  pine 
beams  each  6"  X  16"  with  a  sheet  of  plate-iron  16"  deep  bolted  between 
them.  The  girder  carries  a  load  o/ 13000  pounds  at  its  centre;  of 
whal  thickness  should  the  plate  be  ?  Tlie  girder  supports  a  plastered 
ceiling. 


208  SAFE    BUILDING. 

Strength  Of  From  Table   XIII  we  find  that  a  Georgia  pine 

wooden  part,  beam  G"  X  16"  of  20-foot  spaa  will  safely  carry 
without  cracking  plaster  7980  pounds  uniform  load,  or  3990  pounds 
at  its  centre  (See  Case«(6)  Table  VII,)  so  that  the  two  wooden 
beams  together  carry  7980  pounds  of  the  load,  leaving  a  balance  of 
5020  pounds  for  the  iron  plate  to  carry.  The  deflection  of  a  20-foot 
span  Georgia  pine  beam  6"  X  16"  with  3990  pounds  centre  load  will 
be,  Formula  (40) 

1_     3990.  2408 

°  ~~48'       e.        i. 

e  for  Georgia  pine  (Table  IV)  is  =  1200000  and 

t  =  -j^  (Table  I.  section  No.  2),  or 

6J68_2Q4g  therefore 
*  12 

.        1          3990.        2403       ^,,„ 
0= — 0  47" 

48  •    1200000.  2048  ~' 

3jjg  Qf  We  now  have  a  wrought-iron  plate  which  must 

Iron  Plate,  carry  5020  pounds  centre  load,  of  a  span  of  20  feet, 
16"  deep,  and  must  deflect  under  this  load  only  0,47". 
Inserting  these  values  in  Formula  (40)  we  have : 
1        5020.2403 
0'47=48-    -iT^ 
From  Table  IV  we  have  for  wrought-iron 
g=  27000000 

While  for  i,  we  have  (Table  I.     Section  Xo.  2) 

.      6.d3      6.  163 

i  =  -^-^  =  ^^  =  3Al.b 

Inserting  these  values  and  transposing  we  have : 

5020.      2403 , 

^~  48.  27000000.  341.  0,47"""'^^ 
Or  the  plate   would   have  to  be  ^"  X  16".      Now  to  make   sure 
that  this  deflection  does  not  cause  too  gi-eat  fibre  strains  in  the  iron, 
we  can  calculate  these  from  Formulae  (18)  and  (22).     The  bending 
moment  at  the  centre  will  be  (22) 

5020.  240 

m  = z =  301200 

4 

The  moment  of  resistance  will  be  (Table  I.     Section  No.  2) 

h.d-^      0,33.162 

r=-^r-  = t; =  14 


FLITCH    PLATE    GIKDER. 


209 


And  from  (18) 


=  r,  or  transposing  and  inserting  values, 


(7) 

(^)  =  52"|»2  =  21514  pound.. 


As  the  safe  modulus  of  rupture  of  wrought-iron  is  only  12000 
pounds  (Table  IV)  we  must  increase  the  thickness  of  our  plate. 
Let  us  call  the  plate  |"  X  16",  we  should  then  have 


5.  162 


(7)= 


8.  6 
301200 
2G,(J7 


=  26,67  and 


11256. 


So  that  the  plate  would  be  a  trifle  too  strong.  This  would  mean  that 
both  plate  and  beams  would  deflect  less.  The  exact  amount  niiglit 
be  obtained  by  experimenting,  allowing  the  beams  to  carry  a  little 
less  and  the  plate  a  little  more,  until  their  deflections  were  the  same, 
but  such  a  calculation  would  have  no  practical  value.  We  know 
that  the  deflection  will  be  less  than  0,  47"  and  further,  that  plaster- 
ing would  not  crack,  unless  the  deflection  exceeded  |  of  an  inch 
(Formula  28)  as 

20.0,03  =  0,  6" 
Size  of  Bolts.  In  regard  to  the  bolts,  the  best  position  for  them 
would,  of  course,  be  along  the  neutral  axis,  that  is,  at  half  the  height 
of  the  beam.  For  here  there  would  be  no  strain  on  them.  But  to 
place  them  with  sufiicient  frequency  along  this  line  would  tend  to 
weaken  it  too  much,  encouraging  the  destruction  of  the  beam  from 

oi/no 

C^  ^^  (N 


a 

b 

^ 

t 

i 

♦ 

•f 

4> 

4 

'fe\ 


ti 


1_ 


i- 


,1 


yU^ 


Fig.  137. 


6  i'6 

8 
Fig.    138. 


longitudinal  shearing  along  this  line.  For  this  reason  the  bolts  are 
placed,  alternating,  above  and  below  the  line,  forming  two  lines  of 
bolts,  as  shown  in  Fig.  (137),  The  end  bolts  are  doubled  as  shown; 
the  horizontal  distance,  a-l,  between  two  bolts  should  be  about 
equal  to  the  depth  of  the  beam.     If  we  place  the  bolts  in  /our  exam- 


210  SAKK    lU'ILDING. 

pie,  say  3"  above  and  o"  below  the  neutral  axis,  we  can  readily  cal- 
culate the  size  required.  Take  a  cross-section  of  the  beam  (Fig. 
138)  showing  one  of  the  upper  bolts.  Now  the  fibre  strains  along 
the  upper  edge  of  the  girder,  we  know  are  (  -;  )  or  1200  pounds  per 

sqiuire  inch,  for  the  wood,  and  we  just  found  the  balance  of  the  load 
coming  on  the  iron  would  strain  this  on  the  extreme  upper  edge 
=  1125G  pounds  2"»er  square  inch.  As  the  centre  line  of  the  bolt  is 
only  3"  from  the  neutral  axis  or  f  of  the  distance  from  neutral  axis  to 
the  extreme  upper  fibres,  the  strains  on  the  fdjres  along  this  line  will 
be,  of  course,  on  the  wood  |  of  1200,  or  450  pounds  per  square  inch : 
and  on  the  iron  |  of  112.jG=4221  ])ounds  per  square  inch.  Now, 
supposing  the  bolt  to  be  1"  in  diameter.  It  then  presses  on  each 
side  against  a  surface  of  wood  =  1"  X  6"  or  =  six  stpiare  inches. 
The  fibre  strain  being  450  pounds  jier  s(piare  inch,  the  total  pressure 
on  the  bolt  from  the  wood,  each  side,  is  : 

G.450  =  2700  jjounds. 
On  the  iron  we  have  a  surface  of  1"  X  |"  ^  |  square  inches.  And 
as  the  fibre  strain  at  the  bolt  is  4221,  the  total  strain  on  the  bolt 
from  the  iron  is=|.  4221  =:  2G38  pounds.  Or,  our  bolt  virtually 
becomes  a  beam  of  wrouglit-iron,  circular  and  of  1"  diameter  in 
cross-section,  supported  at  the  points  A  and  B,  which  are  C|"  apart, 
and  loaded  on  its  centre  C  with  a  weight  of  2G38  pounds. 
Therefore  we  have,  at  centre,  bending-moment  (Formula  22) 

2G38.  G| 

in  = T— 2  =  43G9. 

4 

From  Table  I,  Section  Xo.  7,  we  know  that  for  a  circular  section, 
the  moment  of  resistance  is. 


■=n-'=;fa)=«. 


098 

Xow  for  solid  circular  bolts,  and  which  are  acted  on  really  along 

their  whole  length  it  is  customary  to  take  {  -i=  )  the  safe  modulus  of 

rupture  rather  higher  than  for  beams.  Where  the  bolts  or  pins 
have  heads  and  nuts  at  their  ends  firmly  holding  together  the  parts 
acting  across  them  they  are  taken  at  18000  pounds  for  steel  and  at 
15000  pounds  for  iron.  We  have  therefore  transposing  (Formula 
IS)  for  the  required  moment  of  resistance 

43G9  T  .         ,  .  ,        .  .       , 

0,291.     Inserting  this   value   for   r   in   the 


15000' 


felZK    OK    ISOLTS,  2Ji 

abovi;  wu  li;ive  for  tlie  radius  of  bolt, 
-  .-.  r''=0,2;U  and 

^=  Vir  ^''2''i  =  \'/o,3  7o-t 

=  0,718" 
Or  the  diameter  of  liolt  should  be  1,43G"  or  say  1  7-16".  But  1"  will 
be  quite  ample,  as  we  must  remember  that  the  strains  calculated  will 
come  only  on  the  one  bolt  at  the  centre  of  span  of  beam ;  and  that, 
as  the  beam  remains  of  same  cross  section  its  whole  length  the 
extreme  fibre  strains  decrease  rapidly  towards  the  supj)orts,  and 
therefore  also  the  strains  on  the  bolts.  The  end  bolts  are  doubled 
however,  to  resist  the  starting  there  of  a  tendency  to  longitudinal 
shearing.  We  might  further  calculate  the  danger  of  the  bolt  crush- 
ing the  iron  plate  at  its  bearing  against  it;  or  crushing  the  wood 
each  side ;  or  the  danger  of  the  iron  bolt  being  sheared  off  by  the 
iron  plate  between  the  wooden  beams ;  or  the  danger  of  the  iron 
bolt  shearing  off  the  wood  in  front  of  it,  that  is  tearing  its  way  out 
through  the  wood ;  but  the  strains  are  so  small,  that  we  can  readilv 
see  that  none  of  these  dangers  exist. 

Keyed  Girders.  Another  method  of  adding  to  the  sum  of  the 
separate  strengths  of  the  girders  is  to  j)lace  one  under  the  other  mak- 
ing a  straight  joint  between  the  two  parts  and  to  drive  in  hard  wood 
keys,  as  shown  in  Fig.  l-iO. 

The  keys  can  either  be  made  a  trifle  thicker  than  the  holes  and 
the  beams  then  firmly  bolted  together  so  as  to  take  hold  of  keys 
securely;  or,  keys  can  be  shaped  in  two   wedged-shaped  pieces  to 
^  each  key,  and  driven 

into  the  hole  from  op- 
posite sides,  after  the 
beams  are  fi  r  m  1  y 
bolted.  In  the  latter 
case,  care  must  be 
taken  that  the  joint 
Pig,  139,  between   the   opposite 

wedges  is  slanting  or  diagonal,  and  7iot  liorizontal  or  else,  of  course, 
the  keys  would  be  useless.  Either  method  allows,  for  tightening  up, 
after  shrinkage  has  taken  place.  Iron  bands  are  frequently  used 
in  place  of  bolts  but  they  are  more  clumsy,  less  liable  to  all 
fit  exactly,  and  besides  do  not  allow  for  tightening  up  so  easily 
as     with    bolts.        Where    beams    are    very    wide,   however,    the 


I  if: 
r 


212 


SAKK    ISril.DINC.. 


bands    are    -very    advantageous.       Tredgold    says    the   keys    should 

be    twice    as    wiik',    as    high;     and    tliat    tlie    sum    of    all    their 

heights  should    t-ijual  one    and  a  third  times  the    dejith  of    girder. 

They  can  be  easily  calculated,  however.     As  the  main  strain  on  them 

is  a  horizontal  shearin<i    strain,  and  tlie    stress    or 
Horizontal  .        ? 

s:-.earing  on       resistance  to  shearing  is  greatest  across  the    grain 

Keys.  the   keys   should   of   course,  be   placed   with  their 

grain  running  as  nearly  as  possible  A'ertically.      Of  course,  as  the 

•Greatest  horizontal  shearing  exists  near  the  supports,  the  end  wedges 

should  be  the  strongest ;  it  is  customary,  however,  to  make  them  all 

of  the  same  size  for  convenience  of  execution.     Tlie  amount  of  the 

horizontal  shearing  is  found  by  Formula  (13). 

Besides  the  horizontal  shearing  strain  there  will  also  be  a  crushing 
strain  on  the  sides  of  wedges,  which  will  be  greatest,  where  the 
greatest  fibre  strains  exist.  This  of  course,  is  at  the  point  of  the 
greatest  bending  moment  on  the  beam.  Let  us  consider  the  wedge 
Compression  ^t  A-B  Fig.  (140)  which  has  been  drawn  enlarged 
en  Keys,    in  Fig.  (139.) 

The  lower  half  of  the  girder,  being  in  tension,  in  trying  to  stretch 
its  fibres  meets  with  the  resistance  of  the  wedge  along  E  F,  therefore 
tends  to  crush  or  compress  this  surface.  The  amount  of  this  com- 
pression, per  square  inch,  will  be  e(|ual  to  the  average  fibre  strain 
between  E  and  F.  Now  the  fibre  strain  at  A  can  be  readily  found,  by 
finding  the  "bending  moment"  at  A  and  dividing  tliis  by  the  moment 
of  resistance  of  the  girder  (see  Table  I)  and  Formula  (18).     This 


Q2od) 


(4^00) 


Fig.  140. 


gives  the  fibre  strains  at  A.  The  average  fibre  strain  on  E  F  will  be 
to  the  strain  at  A  as  the  distance  of  a;  from  the  neutral  axis,  is  to  the 
depth  of  half  the  beam;  .x being  the  centre  of  E  F,  or  : 

Extreme  fibre  strain  at  A  :  average  fibre  strain  on  Avedge  ^  E  A : 
E  X.  The  amount  of  the  compression  on  E  F  will  of  course  equal 
the  area  of  wedge  at  E  F  (that  is  E  F  multiplied  by  the  breadth  of 


KKYKI)    OIKDKKS.  21.1 

girder),  and  tliis  area  iiiiiltii)lied  by  the  average  fibre  strain  on  !■:  !•'. 
The  greatest  eonipression  on  E  F  will  of  course  be  at  F,  and  ccjual 
to  twice  the  average  iiljre  strain,  as  E  F  =r  2.  E  x. 

In  the  same  way,  we  find  that  the  upper  lialf  of  the  girder,  being 
in  compression,  is  forcing  its  fibres  towards  the  centre  causing  com- 
pression on  the  surface  D  C.  The  amount  of  this  compression  is 
found  similarly  as  for  that  on  E  F,  the  only  difference  being  in  the 
difference  in  bending  moments  at  B  and  A.  The  key  therefore 
becomes  virtually  a  cantilever,  the  built-in  part  being  l)ctween  E  F 
and  C,  and  tlie  load  applied  on  the  free  end  C  I),  the  load  being  a 
uniform  (jue  and  ecjual  to  the  amount  of  tbe  compression  on  C  D. 
Weakest  Point  '''he  Aveakest  point  of  the  girder  itself  will  be  either 
of  Girder.  ,^t  ji^g  poj^jj  „£  gi-^atest  bending  moment,  or  at  key 
nearest  to  it,  where,  of  course  the  girder  will  not  Ijc  of  full  section, 
being  weakened  in  the  part  cut  away  for  key.  An  example  will 
more  fully  illustrate  all  of  the  foregoing. 

Example. 
A  spruce  (jirder  {Fig.  140)  of  •dO-foot  clear  span  is  hullt  up  of  two 
fjirders  10"  X  1-'"  each,  inaJdng  the  whole  section  10"  X  24". 
Georgia  pine  keys  are  used,  each  G"  X  12"  {and,  of  course)  10" 
across  girder;  they  are  placed  w-ith  grain  vertically  3'  4"  hetween 
centres  and  same  distance  from  centre  of  last  key  to  support.  The 
girder  helps  support  a  plastered  ceiling.  What  is  the  safe  centre-  load 
on  girder  ? 

Calculation  of  ^lie  girder  is  (dr=)  24"  deep  and  (L  =  )  thirty 
Keyed  girder,  feet  long  ;  now  l^.  L  would  be  35,  therefore  d  is  less 
than  1^  L,  and  from  rule  contained  in  Table  VIII  for  spruce  we 
must  calculate  for  deflection,  not  rupture,  in  order  to  be  safe. 
Formula  (40)  gives  the  rule  for  deflection  of  a  centre  load  on  a 
girder  or  beam.     It  is  : 

S         1      w.l^ 

48  ■  TTT  "''  t''^"^PO^i»^"' 

O.AS.e.i       ,  ,  ,   ,        ,  p 

"'  =  — -J- —    where  iv  would  be  the  safe  centre  load,  in 

pounds.     Xow  in  order  not  to  crack  plastering,  we  have  from  For- 
nufla  (28) 

S  —  L.  0,03  or 

S  =  30.  0,03 

=     0,1). 
Frou^  Table  IV  we  have  for  spruce: 


214 


SAFE    ISUILOING. 


10" 


^_ 


'KEY  N> 


850000 
From  Table  T,  Section  No.  1 7,  we  have  for'  the 
weakest   section   of   the    girder,  which  would  be 
tliroiigh   a  key,   and    as  shown  in  Fig.  141, 


f  =  ^-0/'-'A^) 


10 


_  {^  .  (--243  —  63)  =  11340,  therefore 

—  12    ^  ^ 

I4i_       inserting  these  values  in  the  transposed  Formula(40) 
0,9.48.8.50000.11340 


'"=  3603 

=  8925 
Or  the  safe  centre  load,  not  to  crack  plastering,  would  1)0,  say  9000 
pounds. 

End  Keys.  Now  let  us  try  the  keys.     We  first  take  the  great- 

est horizontal  shearing,  which  will  be  at  the  end  keys. 

The  vertical  shearing  at  these  keys  will  be  equal  to  the  reaction 
(see  Table  VII,  or  Formula  11.) 

As  the  load  is  central,  each  reaction  will,  of  course,  be  one-half 
the  load,  or  4500  pounds,  therefore  the  vertical  shearing  strain  at 
end  key,  will  be  (a  little  less  than) 

x  =  4500 
Now  from  Formula  (13)  we  know  that  the  horizontal  shearing  strain 
at  the  same  point  is  : 
3       X 

T  ■  Ti 

For  the  area  we  take  the  full  area  of  cross-section  or  0  =  10.24  =  240, 
therefore  horizontal  shearing  strain  : 

■    '        =   28,125     pounds     per     square    inch 


The 


2.  240 

amount  of  this  strain  that  will  act  on  each  key  is,  of  course,  equal  to  the 
area  at  the  neutral  axis  from  centre  to  centre  of  key,  or  40.  10  =  400 
stjuare  inches  multiplied  by  the  strain  per  square  inch,  or 

400.     28,125  =  11250  pounds. 
To  resist  this  we  have  a  key  12"  X  10"  =  120  scpiare  inches,  being 
sheared  across  the  grain.     From  Table  IV  we  know  that  the  safe 
shearing  stress  of  Georgia  pine  across  the  grain  or  fibres,  is  : 

(X\^blO   pounds    so  that  the  key   could  safely  stand  an 

amount  of  horizontal  shearing 

^570.     120  r=  GS400  pounds 


SriEAlUNt;    OS    KKYS.  215 

or  more  than  six  times  the  actual  strain.  Had  we,  liowcvcr,  placed 
the  grain  of  the  key  horizontal,  the  shearing  would  he  with  the  grain 
or  along  the  fibres;  the  safe  shearing  stress  this  way  for  Georgia 
pine  (Table  IV)  is  only  BO  pounds  per  scpiare  inch,  so  that  the  key 
would  only  have  resisted 

=  50.120  ==  (1000  pounds,  or   it  would  have   been    in 
serious  danger  of  splitting  in  two. 

Central  Keys.  Now  take  the  Key  A  B  immediately  to   the  right 

of  the  weight.     The  bending-moment  at  A  will  be  (Taljle  VII) 

m,^=  4500.166  =  747000 
and  at  B 

m„=  4500.154  =  69.3000 
Now  at  A  and  !>,  the  girder  being  uncut,  the  moment  of  resistance 
will  be  (Table  I,  Section  No.  2) 

10.  242 
r=  —. =  960 

0 

Dividing  the  bending  moment  by  the  moment  of  resistance  (Formula 

18  transposed)  gives  the  extreme  fibre  strains, 

.         74  7000 
at  A  =:=  -Q  .„      =r  778  pounds. 

69.3000 
and  at  B  =     ^g^—  =  722  pounds. 

Now  the  centres  (x  and  y,  see  Fig.  139)  of  each  side  of  key  will  be 
1|"  from  neutral  axis,  the  extreme  fibres  being,  of  course,  12"  distant 
from  neutral  axis,  therefore  average  strain  on  side  of  key  at  A.    (Or 

1^ 
X,   Fig.    139)  := -5  .  778  =  97  pounds, 

and  at  B  (or  >j,  Fig.  139)  =  lf  .  722  —  90  jiounds. 

The  extreme  compression,  will,  of  course,  be  on  the  lower  edge  of 

key,    at   A   and   will   be  =  2.97  =  194    pounds   per    square   inch. 

From   Table  IV  we    find    that    Georgia   pine   will   safely   stand    a 

pressure  of  200  pounds  per  scpiare  inch,  across  the  fibres,  so  that  we 

are  just  a  little  inside  of  the  safety  mark.     We  now  have  to  consider 

our  key  as  a  cantilever   with  cross-section  10"  wide  and  12"  deep, 

projecting  3"  beyond  the    support    and  loaded  uniformly   with   a 

weight  e(|ual  to  90  pounds  per  square  inch,  or 

u  =  90.3.10  =  2700  pounds. 

Now  the  bending  moment  at  support  is,  (Formula  25.) 

2700.3 
m  =  — ^ —  =r  4050  pounds. 


216 


SAFE    BUn.DING. 


Till."  inoiiR'nt  of  resistant'o  (Table  T,  Section  2)  is 
10.122 


6 


=  240 


Tliercfore  (Formula  18)  the  extreme  fibre  strains  on  key 

111        4050 

=  y  =  240  =  ^  '  po""'^^- 

Or  not  enough  to  be  even  considered  seriously. 

Another  method  of  conibining  and  strenijthening 

Notched  ,  .  ,1  1  1 

girders,    wooden    girders,   is    to  cut  them   with    saw-siiaped 

notches,  as  shown  in  (Fig.  142)  and  fit  the  teeth  closely  together, 

firmly  bolting  the  two  parts  together,  so  as  to  force  them  to   act 


[J 


Fig.  142. 
together  as  one  girder.  Sometimes  the  top  surface  slants  towards 
each  end,  and  iron  bands  are  driven  on  towards  the  centre,  till  they 
are  tight.  But  bolts  are  more  reliable,  and  not  likely  to  slip  ;  where 
the  gh-der  is  broad,  they  should  be  doubled,  that  is,  placed  in  pairs 
across  the  width  of  girder.  The  distance  between  bolts  should  not 
exceed  twice  the  depth  of  girder.  Great  care  must  be  taken  to  get 
the  right  side  up.i  Many  text-books  even  being  careless  in  this 
matter".  It  must  be  remembered  that  the  upper  fibres  are  in  com- 
pression, crowding  towards  the  centre,  while  the  lower  ones,  in  ten- 
sion, are  pressing  away. 

The  girder  must  therefore  be  placed,  as  shown,  so  that  the  two 
sets  of  fibres  will  meet  at  the  short  joints  and  oppose  each  other. 
The  girder  is  easily  calculated  similarly  to  the  former  example.  The 
crushing  on  C  D  or  A  B  can  be  found,  and  also  the  stress  on  their 
extreme  edges ;  this  must  not  exceed  the  safe  stress  of  the  material 
for  compression  along  the  fibres.  Then  D  B  or  C  A  must  have 
area  sufficient  to  resist  the  horizontal  shearing  strain. 
^^^^^  In  all  these  girders  the   most   careful   fitting   of 

Bearings,  joints  is  necessary;  then,  too,  the  ends  must  have 
sufficient  bearing  not  to  crush  under  the  load.  Thus,  take  the  former 
example,  the  reaction  was  4500  pounds  ;  the  safe  resistance  of  spruce 
to  crushing  across  the  fibres  is  (Table  IV)  =  75  pounds. 

We  need  therefore  an  area  —  ^^  =  60  sejuare  inches,  and  as 
the  girder  is    10"    broad   it    should  bear  on  each   support  f  f  =  6 

1  The  reasons  for  placing  the  notched  girder  in  the  position  shown  in  Figure 
142  have  been  fully  stated  by  me  in  a  letter  published  May  10,  1890,  in  the 
American  Architect,  Vol.  XXYIII,  No.  750. 


CONTIXL'OUS    OIIJDERS. 


m 


inches.  The  end  of  girder  should  Ijc  deep  enough  to  resist  vertieal 
shearing.  In  our  case  it  is  trilling,  and  we  need  not  consider  it.  In 
all  of  these  examples  wo  have  omitted  the  weight  of  the  girder,  to 
avoid  complication.  This  should  really  he  taken  into  account,  iu 
such  a  long  girder,  and  treated  as  an  additional  but  uniform  load. 
Continuous  Wlien  girders  run  over  three  or  more  supports  in 

Girders,  one  piece,  that  is,  are  not  cut  apart  or  jointed  over 
the  supports,  the  existing  strains  and  reactions  of  ordinary  girders, 
are  very  much  altered.  These  are  known  as  "continuous  girders." 
If  Tve  have  (Fig.  143)  three  supports,  and  run  a  continuous  girder 
over  them  in  one  piece  and  load  the  girder  on  each  side  it  will  act  as 


shown  iu  Fig.  143 ;  if  the  girder  is  cut  it  will  act  as  shown  in  Fig. 
144.  Very  little  thought  will  show  that  the  fibres  at  A  not  being 
able  to  separate  in  the  first  case,  though  they  Avant  to,  nnist  cause 
considerable  tension  in  the  upper  fibres  at  A.  This  tension,  of 
course,  takes  up  or  counterbalances  part  of  the  compression  existing 
there,  and  the  result  is  that  the  first  or  continuous  girder  (Fig.  143) 
is  considerably  stronger,  that  is,  it  is  less  strained  and  considerably 
stiff er,  than  the  sectional  or  jointed  girder  (Fig.   144).     Again  we 


Fig.  144. 

can  readily  see  that  the  great  tension  and  conflict  of  the  opposite 
stresses  at  A  would  tend  to  cause  more  pressure  on  the  central  post 
in  Fi"".  143,  than  on  the  central  post  in  Fig.  144,  and  this,  in  fact, 
is  the  case. 


218 


SAFE    BUILDING. 


--»         CO 

~" 

1 

^ 

r-N 

S         i. 

S        ""                    K        —             1 

c 

s" 

c      o 

^  ./ 

^ 

+ 

5-2 

S^T 

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-M 

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S 

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+ 

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c 

g 

c 

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8 

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3  reaction. 

reaction, 
to..  —  3  .  n\ 

2 
•§ 

i     .2 

+ 

2      + 

2 

1— ( 
> 

X 

a 

3 
o 

s 

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a. 

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11 

pq 

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tn     o     m            S    "^J 
1^1          VA    II 

M      O      O 

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3  ~- 

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o 

Two  equa 
carrying 
load   but 
equal. 

Two  ccpia 
carrying 
load  = 
equal. 

O    J 

1 

(D- 

rrr-0 

^ 

3 

J 

p-© 

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a 

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1 

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5 

Where  f5,  (5,,  ^11  =  the  amount  of  deflection  in  inches,  if  girder  of  uniform  cross- 
seciiou  tliiougbout. 
"        e=:the  raoilulus  of  elasticitv  of  the  material,   in  pounds-inch,  (see 

Table  IV). 
''        i  =  the  moment  of  inertia  of  the  cross-section,  in  inc'-ies,  (see  Table  I). 


CONTINUOUS   GlItDKRS. 


219 


/-N 

'T: 

^ 

S             S     a, 

j;> 

C3 

~      F^ 

a" 

+ 

^       +  -:: 

+ 

gi 

+   4 

+ 

1       f  ^ 

s" 

"P 

^ 

^- 

3           '^ 

+ 

V 

e       •-      L.  '^ 

■-     w      ~ 

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s 

n 

o 

Ss. 

w 

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— < 

^ 

It 

<B 

Where  w,  Wi,  iCn  =  central  concentrated  loads  in  pounds,  on  either  span,  being 
equal,  when  so  stated. 
"        I,  It,  /„  =  the  length  of  respective  spans,  in  inches,  all  being  equal. 
"        u,  i(,,?t.i=  uniform  loads  on  each  span,  in  pounds,  all  being  equal. 
"       }),  r,  .<!,(7,  r=the  amount  of  respective  reactions.  In  pounds. 
"       m  =  the  tending  moment,  in  pounds-inch. 


220 


SAFE   BUILDIXG. 


CO 

1 

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g  |(N          S  l^-' 

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1  "    H 

o 
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a 

a 

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6            G 

3 

s" 

-^3         "S 

-3^5   'A^ 

S 
o 

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til  one  load  r=  i 
(not  central,) 
at  any  point 

A  B  <  BC 

Trussed  Beam 
ith  uniform  loa 

=  u 

1  one  central  str 

A  B  =  B  C 

S  -  "■  =  1  11 

?    =    1  :=    ?    11 

> 

^                     1 

>        5 

"        1 

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M 

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r^-e 

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1 

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i 

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= 

/ 

/ 

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y- 

1 

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id_H9 

4 

1 

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V-€) 

IJiUSSKD    UK  A  MS. 


221 


t, 

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t. 

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^'''"''l"J 

222  SAi'i;  I'.uir.Dixd. 

Ill  Tabic  XVIT,  pages  218  and  210,  arc  given  tlie  various  formui;e 
for  reactions,  greatest  bending  momiMits  and  deflections,  for  the  most 
usual  cases  of  continuous  girders.  The  architect  can,  if  he  wishes, 
neglect  to  allow  for  the  additional  strength  and  stiffness  of  continu- 
ous girders,  as  both  are  on  the  safe  side.  But  he  must  never  ovei-- 
Variation  of  look  the  fact  tJiat  the  central  reactions  are  much 
Reactions!  {greater,  or  in  other  words,  that  the  end  supports 
carry  less,  and  the  central  supports  carry  more,  than  Avhen  the  girders 
are  cut. 

Bending  moments  can  l)e  figured,  at  any  desired  point  along  a 
continuous  girder,  as  usual,  subtracting  from  the  sum  of  the  reactions 
on  one  side  multiplied  by  their  respective  distances  from  the  point, 
the  sum  of  all  the  weights  on  the  same  side,  multiplied  by  their  re- 
spective distances  from  the  point.  Sometimes  the  result  will  he 
negative,  which  means  a  reversal  of  the  usual  stresses  and  strains. 
Otherwise  the  rules  and  formula?  hold  good,  the  same  as  for  other 
girders  or  beams.  Table  XYII  gives  all  necessary  information  at 
a  glance. 

Strength  is  frequently  added  to  a  girder  or  beam  by  trussing  it,  as 
shown  in  Table  XVIII,  pages  220  and  221.  One  or  two  struts 
are  placed  against  the  lower  edge  of  a  beam  and  a  rod  passed 
over  them  and  secured  to  each  end  of  the  beam ;  by  stretching  this 
rod  the  beam  becomes  the  compression  chord  of  a  truss  and  also  a 
Trussed  continuous  girder  running  over  one  or  tAvo  supports. 

Beams.  There  must  therefore  be  enough  material  in  the  beam 
to  stand  the  compression,  and  in  addition  to  this  enough  to  stand  the 
transverse  strains  on  the  continuous  girder.  If  the  loads  are  concen- 
trated immediately  over  the  braces,  there  will  be  no  transverse  strain 
wdiatever,  but  the  braces  will  be  compressed  the  full  amount  of  the 
resj)ective  loads  on  each.  In  the  case  of  uniform  loads,  transverse 
strains  cannot  be  avoided,  of  course,  but  where  loads  are  concen- 
Struts  placed       trated   the  struts  should  always  be  placed  immed- 

under  load.  iately  under  them.  Even  where  loads  are  placed 
very  unevenly,  it  is  better  to  have  the  panels  of  the  truss  irregular, 
thus  avoiding  cross  or  transverse  strains.  This  same  rule  holds  good 
in  designing  trusses  of  any  kind. 

Necessary  Table  XVIII  shows  very  clearly  the  amount  and 

Conditions,  kind  of  strains  in  each  part  of  trussed  beams.  Where 
there  are  two  struts  and  they  are  of  any  length  care  must  be  taken 
by  diagonal  braces  or  otherwise,  to  keep  the  lower  ends  of  braces 
from  tipping  towards  each  other.     Theoretically  they  cannot  tip,  but 


TKUSSKl)     ItKAMS.  223 

practically,  sometimes,  llicy  do.  Can-  must  be  taken  that  the  heam 
is  braced  sideways,  or  else  it  must  lie  figured  fur  its  safety  against 
lateral  flexure  (Formula  5.)  'J'lien  it  must  have  material  enough  not 
to  shear  off  at  supports,  nor  to  crush  its  under  side  where  lying  on 
support.  The  ends  of  rods  must  have  suflicient  bearing  not  to  crush 
the  wood.  Iron  shoes  are  sometimes  used,  l)ut  if  very  large  are  apt 
to  rot  the  wood.  In  that  case  it  is  well  to  have  a  few  small  holes  in 
the  shoes,  to  allow  ventilation  to  end  of  timber.  If  iron  sti-aps  and 
bolts  are  used  at  the  end,  care  must  be  taken  that  the  strap  does  not 
tear  apart  at  bolt  holes ;  that  it  does  not  crush  itself  against  bolts ; 
that  it  does  not  shear  off  the  bolts,  and  that  it  does  not  crush  in  the 
end  of  timber.  Care  must  also  be  taken  to  have  enough  bolts,  so  that 
they  do  not  crush  the  wood  before  them,  and  to  keep  the  bolts  frora 
shearino-  out,  that  is  tearing  out  the  wood  before  them.  In  all  truss- 
es and  trussed  works  the  ioints  must  be  carefully 
liTiporx^ncQ  •' 

of  Joints,  designed  to  cover  all  these  points.  Many  architects 
give  tremendous  sizes  for  timbers  and  rods  in  trusses,  thus  adding 
unnecessary  weight,  but  when  it  comes  to  the  joint,  they  overlook  it, 
and  then  are  surprised  when  the  truss  gives  out.  The  next  time 
they  add  more  timi)er  and  more  iron,  till  they  learn  the  lesson.  It 
must  be  remembered  that  the  strength  of  a  truss  is  only  e(|ual  to  the 
streno-th  of  its  weakest  part,  be  that  part  a  member  or  only  a  part  of 
a   joint.     This  subject  will  be  fully  dealt  w^ith  in   the   cha[)ter  on 

Trusses. 

The  deeper  the  truss  is  made,  that  is,  the  further 
Depth  '■ 

Desirable,   we  separate  the  top  and  bottom  chords,  the  stronger 

will  it  be ;  besides  additional  depfh  adds  very  much  to  the  stiffness 

of  a  truss. 

_   „     ^.  All  trussed  beams,  and  all  trusses  should  be  "cam- 

Deflection 

of  Girders  bered  up,"  that  is,  built  up  above  their  natural  lines 
and  Beams,  sufliciently  to  allow  for  settling  back  into  their  cor- 
rect lines,  when  loaded.  The  amount  of  the  camber  should  equal  the 
calculated  deflection.  For  all  beams,  girders,  etc.,  of  uniform  cross- 
section  throughout,  the  deflection  can  be  calculated  from  FormuliB 
(37)  to  (42)  according  to  the  manner  of  loading.  For  wrought-iron 
beams  and  plate-girders  of  uniform  cross-section  throughout,  the  de- 
flection can  be  calculated  from  the  same  formulae ;  where,  however, 
the  load  is  uniform  and  it  is  desired  to  simplify  the  calculation,  the 
deflection  can  be  cpute  closely  ealculated  from  the  following  Fornuda: 
Uniform  Cross-  ^  .   -^ '  Ci^^ 


section  and 
Load. 


,(/ 


224  SAFK    BUILDING. 

Where  3=^  the  greatest  tlelleeticin  ;it  centre,  in  inches,  of  a 
■wrou"-ht-iron  beam  or  ])Uite  girder  of  uniform  cross-section  through- 
out, and  carrying  its  total  safe  uniform  load,  calculated  for  rupture 
only. 

Where  L  =  the  length  of  span,  in  feet. 

"\Vhei-e  r/  =  the  total  depth  of  beam  or  girder  in  inches. 

If  beam  or  plate  girder  is  of  steel,  use  64^  instead  of  75. 

If  the  load  is  not  uniform,  change  the  result,  as  jarovided  in  cases 
(1)  to  (8),  Table  YII. 

For  a  centi-e  load  we  should  use  93 1  in  jalace  of  75  or 

Uniform  Cross-  5;  L'^  ^oa\ 

section,  Centre      "  —  ;,...;    ,/  ^^^^ 

Load.  ^'"'"f " 

Where  values  are  the  same,  as  for  Formula  (79)  except  that  beam 
or  girder  carries  its  total  safe  centre  load,  calculated  for  rupture  only. 
If  beam  or  girder  is  of  steel  use  80§  instead  of  93|. 
Therefore  not  to  crack  plastering  and  yet  to  carry  their  full  safe 
loads,  wrought-iron  beams  or  plate  girders  should  never  exceed  in 
length  (measured   in  feet)  twice    and   a   quarter    times   the  depth 
(measured  in  inches),  if  the  load  is  uniform,  or 
Safe  length, 
uniform  Cross-       2^.d  =  L  (81) 

section  and  "  i' 

Load. 
Where  L  =  the  ultimate  length  of  span  (not  to  crack  plastering), 

in  feet,  of  a  wrought-iron  beam  or  plate  girder,  of  uniform  cross- 
section  throughout  and  uniformly  loaded  with  its  total  safe  load. 

Where  rf^the  total  depth  of  beam  or  girder  in  inches. 

If  beam  or  girder  is  of  steel,  use  2  instead  of  2|. 

If  the  load  is  central  the  length  in  feet  should  not  exceed  2 1  times 
the  depth  in  inches,  or 
Safe  length, 
uniform  Cross-       9  4    ^ -— /^  (82) 

section,  Centre       -  5"  •  ^     -' 

Load. 

Where  X  =  the  ultimate  length  of  span  in  feet  (not  to  crack 
plastering),  of  a  wrought-iron  beam  or  plate  girder,  of  uniform  cross- 
section  throughout,  and  loaded  at  its  centre  with  its  total  safe  load. 

Where  d  =  the  total  depth  of  beam  or  girder  in  inches. 

If  beam  or  girder  is  of  steel  use  2-|  instead  of  2|. 
Deepest  beam         One  thing  should  always  be   remembered,  when 

eccfnomical.  "^ing  iron  beams,  and  that  is,  that  the  deepest  beam 
is  always  not  only  the  stiffest,  but  the  most  economical.  For  instance, 
if  we  find  it  necessarv  to  use  a  loi"  beam  — 105  pounds  per  yard, 


DEKLFXTIOX    OF    TKUSSES.  'i'lo 

it  will  be  cheaper  to  use  instead  the  12"  beam —  96  pounds  per  yard. 

The  latter  beam  not  only  weighs  9  pounds  per  yard  less,  but  it  will 

carry  more,  and  deflect  less,  owing  to  its  extra  two  inches  of  depth. 

This  same  rule  holds  good  for  nearly  all  sections. 

^  „     ^.  To  obtain  the    deflections  of   trussed   beams   or 

Deflection 

of  Trusses,  girders  by  the  rules  already  given  would  be  very 

complicated.  For  these  cases,  however,  Box  gives  an  approximate 
rule,  which  answers  every  purpose.  lie  calculates  the  amount  of 
extension  in  the  tension  (usually  the  lower)  chord,  and  the 
amount  of  contraction  in  the  compression  (usually  the  upper) 
chord,  due  to  the  strains  in  each,  and  from  these,  obtains  the  de- 
flections. Of  course  the  average  strain  in  each  chord  must  be  taken 
and  not  the  greatest  strain  at  any  one  point  in  either.  In  a  truss, 
where  each  part  is  proportioned  in  size  to  resist  exactly  the  com- 
pressive or  tensional  strain  on  the  part,  every  part  will,  of  course,  bo 

strained  alike  ;  the  strain  in  the  comi)rcssive    member  being  =  (    ,,  j 

per  square  inch,  throughout  the  whole  length,  and  in  the  tension 

member  ^=  I— A  per  square  inch,  throughout  the  whole  length. 

The  same  holds  good  practically  for  plate  girders,  where  the  top 
and  bottom  flanges  are  diminished  towards  the  ends,  in  proportion  to 
the  bending  moment.  But  where,  as  in  wrought-iron  beams  (and  in 
many  trusses),  the  flanges  are  made,  for  the  sake  of  convenience,  of 
uniform  cross-section  throughout  their  entire  length,  the  "  average  " 
strain  will,  of  course,  be  much  less,  and  consequently  the  beam  or 

girder  stiffer. 

-..     •  If  we  construct  the  graphical  representation  of  the 

Average  strain  o     i  •■ 

in  Chords,  bending  moments  at  each  point  of  beam  (as  will  be 

explained  in  the  next  Chapter)  and  divide  the  area  of  this  figure  in 

inch-pounds  by  the  length  of  span   in   inches,  we  will   obtain   the 

average  strain  in  either  flange,  provided  the  flange  is   of   uniform 

cross-section  throughout,  or 

Uniform  Cross-         f  =  iL  (83) 

section.  I 

Where  v  =  the  average  strain,  in  pounds,  on  top  or  bottom  flange 
or  chord,  where  beam  or  girder  is  of  uniform  cross-section  through- 
out. 

Where  I  =  the  length  of  span,  in  inches. 

A\  here  a  =  the  area  in  pounds-inch  of  the  graphical  figure 
giving  the  bending  mcment  at  all  points  of  beam. 


226  SAFE   BUILDING. 

To  obtain  the  dimensions  of  this  figure  measure  its  base  line  (or 
horizontal  measurement)  in  inches,  and  its  lieight  (or  vertical  meas 
urement)  in  pounds,  assuming  the  greatest  vertical  measurement  as 

r=  (  -L  )  or  =  (  -^  ),  in  i  oiinds,  according  to  which  flange  we  are  ex- 
amining. 

Thus,  in  the  case  of  a  uniform  load,  this  figure  -would  be  a  parabola, 
with  a  base  of  length  equal  to  the  span  measured  in  inches,  and  a 
height  equal  to  the  greatest  fibre  strains  in  pounds ;  the  average 
strain  therefore  in  the  compression  member  of  a  beam,  girder  or 
truss,  of  uniform  cross-section  throughout  would  be,  —  (remembering 
that  the  area  of  a  parabola  is  ec[ual  to  two-thirds  of  the  product  of 
its  height  into  its  base). 

Uniform  load  ''    /  (  "T"  ) 

and  Cross-  i!^l_AZ__or 

section!  I 

(84) 


|-(f) 


J 

"Where  v  =  the  average  strain,  in  pounds,  in  compresf^ion  flange 
or  chord  of  a  beam,  gh-der  or  truss  of  uniform  cross-section  through- 
out and  carrying  its  total  safe  uniform  load. 

"Where  ( -^  )  ^  the  safe  resistance  to  compression  per  square  inch 

of  the  material. 

It  is  supposed,  of  course,  that  at  the  point  of  greatest  bending 
moment  —  or  where  the  greatest  compression  strain  exists  —  that 
the  part  is  designed  to  resist  or  exert  a  stress  =  (  —  j  per  square 
inch.  If  the  greatest  compression  stress  is  less,  insert  its  value  in 
place  oi  (^\     Of  course,  it  must  never  be  greater  than  (-^V 

Similarly  we  should  have 


Uniform  Load 
and  Cross- 
section. 


-f  (7) 


"Where  v  =  the  average  strain,  in  pounds,  in  tension  flange  or 
chord  of  a  beam,  girder  or  truss  of  uniform  cross-section  throughout, 
and  carrying  its  total  safe  uniform  load. 

Where  ( —^  )  =  the  safe  resistance  to  tension,  per  square  inch,  of 

the  material. 

It  being  understood  that  at  the  point  of  greatest  bending  moment 
—  or  where  the  greatest  tension  strain  exists  —  that  the  part  is  de- 


Centre  Load 
Uniform 
Cross-Section. 


(86) 


(8  7) 


DKKI.KCTION,    KOUML'LJ:.  227 

signed  to  resist  or  exert  a  stress  =  ( y)  per  s.iiiare  indi.  If  tliis 
greatest  tensional  stress  is  less  than  {-L^  insert  its  value  in  its  j.jaee 
in  Formula  (85).     Of  eonrsL-,  it  must  never  be  greater  than  (  i  Y 

For  a  beam,  girder  or  truss  with  a  load  eoncentrated  at  the  cen- 
tre, but  with  flanges  or  chords  of  uniform  cross-section  throughout, 
the  average  strain  would  be  just  one-half  that  at  the  centre  ;  for,  the 
bending-moment  graphical-figure  will  be  a  triangle,  and  inserting  the 
values  in  Formula  (83)  would  give  for  the  compression  member  ■" 

-■(7) 

and  for  the  tension  member  : 

■■='■(7) 

The  meaning  of  letters  being  the  same  as  in  Formula?  (84)  and 
(85),  but  the  total  safe  load  being  concentrated  at  the  centre  instead 
of  uniformly  distributed. 

To  obtain  the  amount  of   contraction  or  expansion  due    to    this 

average  strain,  use  the  following  Formula : 

Expansion  or  v.  I 

Contraction  X^ —  (^gg) 

from  Strain.  ^ 

Where  ?;  r=  the  avercuje  strain,  in  jiounds  ^ler  square  inch,  in  eith- 
er chord  or  flange. 

Where  I  =  the  length  of  span,  in  inches. 

Where  e  =  the  modulus  of  elasticity  of  the  material,  in  jjounds- 
inch. 

A\  here  x^the  total  amount  of  extension  or  contraction,  in  inches, 
of  the  chord  or  flange. 

Now  let  us  apply  the  above  rules  to  beams,  plate  girders,  and 
trussed  beams.  Taking  the  case  of  a  beam  or  plate  girder  or  truss 
with  parallel  flanges  or  chords. 

Figure  145  shows  the 
same,  after  the  deflection 
has  taken  place.  We  can 
now  assume  approximate-  H 
ly,  that  C  A  is  equal  to  one- 
half  the  difference  between 

the  contraction  of  G  C  and  '^'^"  ^'^^' 

the  elongation  of  H  D,  or,  what  amounts  to  the  same  thing,  that  CA 


228  SAKK    HIILDIXG. 

is  c:(jnal  to  oiu'-lialf  the  sum  of   tho  contraction  of  the  one  and   the 
elongation  of  tlie  other. 

Further,  we  can  assume  tliat  apjiroximately,  A  B.=^  d  or  the  depth 

of  beam,  and  C  D  =  -^  or  one  half  ihe  span. 

The  curve  C  E  C  will  approximate  a  parabola,  so  that  if  we  draw 

D  F 

a  tanijcnt  C  F  to  the  same  at  C,  we  know  that  D  E  =  E  F^ — ^ — 

or  J)  i''=2.  D  E.     But  as  DE  represents  the  deflection  (  8  )    of 

the  beam,  we  have 

D  F=  2.  S 
Xow  as  C  F  is  normal  to  C  B,  and  C  D  normal  to  .4   B,  we  know- 
that  angles  D  C  F  =  A  B  C;  further,  as  both  triangles  are  right 
angle  triangles,  we  know  that  they  are  similar,  therefore : 

D  F:  C  A.:D  C:  A  B,  or 

2.  8  :   C  A  .-.J.  :  d  or 

S       ^'2        CA.l 
"  —    2.d     —   4.d. 

If  now  we  assume  the  sum  of  the  extension  and  contraction  of  the 
two  flanges  or  chords  to  be  =  x. 

We  have  C  A  ^  —  or 

■2 

Deflection  of  Par-  o        x    I 

allel    Flanges  0=^  (89) 

orChords,  any  8.  a 

Cross-section. 

"Where  8  =  the  deflection,  in  inches,  of   a  beam,  plate  girder  or 
truss,  with  parallel  flanges  or  chords. 

Where  x  =  the  sum  of  the  amount  of  extension  in  tension  chord, 
plus  the  amount  of  contraction  in  compression  chord. 

Where  I  =  the  length  of  span,  in  inches. 

"\Miere  d  =  the  total  depth  of  beam,  girder  or  truss  in  inches. 

Take  the  case  of  a  wrought-iron  plate  girder  or  beam  of  uniform 
cross-section  throughout  carrying  its  full  uniform  load,  we  should 
have  the  strain  at  the  centre  on  the  extreme  fibres  =  12000  pounds 
per  square  inch.  Now  the  average  strain  on  both  upper  and  lower 
flanges  would  be.  Formula  (84)  and  (85). 
v  =  §.  1 2000  =  8000  pounds 
per  s(juarc  inch.     Therefore  amount  of  contraction  in  upper  flange 


DKKLKCTION    CONTINIED.  229 

Formula  (SS),  (and  reincmberin;j;  that,  from  Table  IV,  c  —  27000000) 
__8000.  ^    _     I 
^UOUOOO       3375 
The  elongation  of  the  bottom  (lange  Avould  be  an  ecjual  amount, 
therefore  the  sum  of  the  two 

''  I 
.r,  =  2.x=    "      . 
3375 
_       I 
1G87,5 
Inserting  these  values  in  Fornuila  (S'J)  we  have  the  deflection 

8-       I'       -     I' 

8.1G87, 5.^       135U0.(i 
and  inserting  for  /-=:144.  Z^,  we  have 
^_   144.  Z'-' 
1350U.  (/. 

~ 93f .  d. 
Had  we  assumed  that  the  area  of  flanges  or  chords  diminished  to- 
wards the  suiaports  in  proportion  to  the  bending  moment  or  actual 
stresses  required,  the  average  strain  would,  of  course,  be  12000 
pounds  per  square  inch  throughout  the  entire  length,  no  matter  how 
the  load  might  be  applied. 

Inserting  this  value  in  Formula  (88)  we  should  have  had,  for  the 
amount  of  contraction  of  top  flange 
,_    12')00.  Z   _     / 
27  )U0O00         2250 
The  same  for  the  extension  of  bottom  chord,  or 

^'    ■   ""2250        1125 
Inserting  this  in  Formula  (89)  we  have  for  the  deflection  : 

8-     I'     -    I' 
8.1125.  (/      yooo.  (/ 

Inscrtii.g  144  L'^r^l'^  wq  have 

S       144.Z2 

OOUO.  (/ 

Parallel  flanges 
orChords,  Dl-  5  j^2  (90) 

minished  Cross-  ''  — 

section,  any 
loads. 

"Where  8  =  the  greatest  deflection,  in  iiuhes,  of   a  wrought-iron 

plate  girder,  or  wrought-iron  truss,  with  i)arallel  flanges  or  chords, 

and  where  the  areas  of  flanges  or  chords  are  gradually  diminished 


230 


SAFE    HUILDIXG. 


towards  supports,  and  no  matter  how  the  load  is  applit'd ;  in  no  part 
however  must  the  stresses,  per  scpiare  inch  exceed  respectively  either 

Where  L  =  the  length  of  span,  in  feet. 

Where  f/=r  the  total  depth  (height)  in  inches,  from    top  of   top 
■flange  or  chord  to  bottom  of  bottom  flange  or  chord. 

If  girder  or  truss  is  of  steel,  use  53§  instead  of  62^. 

From  Formula  (90)  and  Formula  (28)  we  get  the  rule  that  (no  mat- 
ter how  the  load  is  applied)  if  we  want  to  carry  the  full  safe  load 
and  not  have  deflection  enough  to  crack  plastering  the  length  in  feet 
must  not  exceed  1|  times  the  total  depth  in  inches. 
For: 

^-    ''''=    G^°^ 

Z  =  621.  0,03.  (Z 

=r 1,875.  d  or  say 

Safe  Length,  Di- 
minished Cross- 
section,  any  L  =  lLd  (91) 
Load,  Parallel                                          ^  ^      ^ 
Flanges  or 
Chords. 

Where  Z  =  the  length,  in  feet,  of  a  wrought-iron  plate  girder  or 

wrought-iron  truss,  with  parallel  flanges  or  chords  and  with  area  of 

flanges    or    chords  diminishing  gradually  towards  supports  and  no 

matter  how  the  load  is  applied;  in  no  part  however  must  the  stresses, 

per    square   inch,  exceed    respectively    either  (^—rj  or  ^ —^j. 

Where  f/  =  the  total  depth  (height),  in  inches,  from  top  of  top 
flange  or  chord  to  bottom  of  bottom  flange  or  chord. 

If  girder  or  truss  is  of  steel,  use  1  f  instead  of  1|. 

We  see  therefore  that  a  beam  of  diminishing  cross-section  through- 
out is  only  about  |  as  stiff,  as  one  with  uniform  cross-section,  as  its 
amount  of  deflection  will  be 
one-half  more  than  that  of  the 
latter.  Both  deflections  are 
approximate  only,  however, 
as  we  see  by  comparing  the 
amount  for  the  uniform  cross- 
section  to  that  obtained  from 
Formula  (79).  The  deflection 
for  varvina;  cross-sections  how- 


SAFK    LKNGTII.  231 

ever  can  be  assumed  as  nearly  enough  correct,  as  these  are  never 
diniinislied  so  nuich  practically  as  we  have  assumed  in  theory.  Xow 
taking  the  case  of  a  trussed  beam. 

Deflection  ^^  Figure  140,  let  A  B  be  one  half  of  a  trussed 

Trussed  Beam,  b^am,  let  B  C  be  the  strut  and  A  C  the  tie.  We 
•will  consider  the  load  concentrated  at  B.  Xow  tlie  first  effect  is  to 
shorten  A  B  by  compression,  let  us  say  to  D  B. 

Then,  of  course,  A  D  will  represent  one  half  of  the  contraction  in 
the  whole  beam  A  G.  Now  the  end  of  rod  .1  moving  to  J)  will,  of 
course,  let  the  point  C  down  to  i?,  if  we  make  D  E  =  A  C. 

But  there  will  be  an  elongation  m  D  E  besides,  due  to  the  tension 
in  it,  Avhich  will  let  it  down  still  further,  say  to  F,  if  D  F=  A  C  -\- 
elongation  in  A  C,  of  course  the  point  B  will  move  down  too,  but  we 
can  overlook  this  to  avoid  complication.  We  now  have  C  F  repre- 
senting the  amount  of  the  deflection.  To  this  should  be  added  the 
amount  of  contraction  of  B  C  due  to  the  compression  in  it.  We  can 
readily  find  C  F. 

We  know  that 


BF=y^)F^"-^^' 

Now  D  F  we  know  is  =  .4  C  plus  the  elongation  oi  A  C  due  to  the 
tension  in  it,  which  we  can  find  from  Formula  (88).  From  same  for- 
mula we  find  the  amount  of  contraction  m  A  G  oi  which  ^  D  is  one- 
half,  subtracting  this  from  A  B  or  —leaves,  of  course,  D  B. 

Now  having  found  B  F  we  substract  from  it  B  C,  the  length  of 
which  is  known,  and  the  balance  is  of  course  the  deflection  C  F ;  to 
this  we  add  the  contraction  of  B  C  and  obtain  the  total  deflection  of 
the  whole  trussed  beam. 

If  the  load  had  been  a  uniform  load,  instead  of  a  concentrated  one 
over  the  strut,  there  would  be  a  deflection  in  that  part  oi  A  G  which 
would  be  acting  as  a  continuous  girder.  But  this  deflection  would 
take  place  between  B  and  G  and  between  B  and  A  and  would  not  af- 
fect the  deflection  of  the  -whole  trussed  beam. 

An  example  will  make  much  of  the  foregoing  more  clear. 

Example. 

Trussed  Beam.        A  trussed  Georgia  Pine  beam  is  16"    deep  and  of 
2i  feet  dear  span;   it  bears  IC"  on  each  support  and  is  trussed  as 


232 


SAFE    BUILDING. 


shown  in  Figure  147.  The 
^    beam  carries  a  uniformly 

distributed  load  of  40800 
'is  pounds  on  the  whole  span 
r    including  weight  of  beam 

and  trussing.     Of  what  size 

should  the  parts  be  ? 


Fig.  147. 


We  draw  the  longitudinal  neutral  axes  of  each  part,  namely  A  B, 
B  C  and  A  C.  The  latter  is  so  drawn  that  the  neutral  axis  of  the 
reaction,  which  is  of  course  half  way  between  end  of  girder  and  E 
(or  8"  from  E)  will  also  jiass  through  A. 

In  designing  trusses  this  should  always  be  borne  in  mind,  that  so 

far  as  possible  all  the  neutral  axes  at  each  joint  should  go  through 

the  same  point. 

^  .  The  beam  A  F  virtually  becomes    a   continuous 

Cross-strains  e         r 

in  Beams.       (birder,  of  two  equal  spans  of  12  feet  or  144  '  each, 

uniformly  loaded  with  20400  pounds  each,  and  supported  at  three 

points  A,  B  and  F.     From  table  XVII  we  know  that  the  greatest 

bending  moment  is  at  B  and 

_tU_20400^_3,_200  pounds-inch. 

8  8  ^ 

The  modulus  of  rupture  for  Georgia  pine  (Table  IV)  is 

^       ^  -200,  therefore  moment  of  resistance  (r)  from  Formula  (18) 


(|)  =  .K 


or 


and  Table  I,  section  No.  2, 

__?;.(/ 2 _  367200 
'  ~~  13  1 200 

b.d -=1S3G 

Xow  we  know  that  fZ=  IG,  or  d'^: 
1836 


256,  therefore 


b:=—^  =  7,2  or    sav  we  need  abeam  71"  x  16"  for  the 
256  '  * 


KXAMPI.K,    TUUSSKD    liEAM.  233 

transverse  strain.     We  must  add  to  this  however  for  the  additional 
conipression  due  to  the  trussing. 

Compression  "^''^  amount  of  the  load  carried  by  strut  C  B,  see 

in  Strut.        Ta])le  XVII,  is 

=  |.  u  from  each  side,  or 

=  25500  on  the  strut  B  C,  of  Avhich 

=  12750  from  each  side. 
If  now  we  make  at  any  scale  a  vertical  line  b  c=zhalf  the  load 
Compression  carried  at  j^oint  B  or  =  12750  in  our  case,  and 
in  Beam.  draw  b  a  horizontally  and  a  c  parallel  to  A  C,  we 
find  the  strain  in  B  A  by  measuring  b  «  =  (32300  pounds)  or  in  A  C 
by  measuring  a  c=:(34G38  pounds)  both  measured  at  same  scale 
as  h  c.  We  find,  further,  in  passing  around  the  triangle  c  b  a  c  — 
(c  b  being  the  direction  of  the  reaction  at  A'),  that  b  a  is  pushing  to- 
wards A,  therefore  compression ;  and  that  a  c  is  pulling  away  from 
.1,  therefore  tension.  Using  the  usual  signs  of -(-for  compression, 
and  —  for  tension,  we  have  then : 

J. -B  =  -[- 32300  pounds, 

A  C=  —  34G38  pounds. 

BC  =  -{-  25500  pounds. 
Had  we  used  Table  XVIII  we  should  have  had  the  same  result 
for : 

Compression  in  A  ^=  ?i^|^,  ^^  =  -4-  32300  pounds  and 
2        Z>  C 

rn       ■       •       i   n       25500  A  C  „.„.,„ 

lension  in  A  C^ — - — -jT-r,^= — ^34038  pounds. 
2       x>  C 

Now  the  safe  resistance  of  Georgia  jiine  to  compression  along  fibres 
(Table  IV)  is 

(  -7;  j=  750  2)uunds. 

If  A  B  were  very  long,  or  the  beam  very  shallow  or  very  thin,  we 
should  still  further  reduce  {-^,  j  by  using  Formulae  (3),  or  (5).     But 

we  can  readily  see  that  the  beam  will  not  bend  much  by  vertical 

flexure  due  to  compression,  nor  will  it  deflect  laterally  very  much,  so 

we  can  safely  allow  the  maximum  safe  stress  per  square  inch,  or  750 

pounds,  that  is,  consider  A  B  a,  short  column. 

The  necessary  area  to  resist  the  compression,  Formula  (2)  is : 

32300  =  0.  750  or 

32300 
a  =  — .—  —  ^43  sciuare  inches. 
7y0  ^ 


234  SAFE   BUILDING. 

As  the  beam  is  IG"  dofp,  this  would  mean  an  additional  thickness 

4  3  —  o  1  i_ 

IB' 16 

Adding  tliis  to  the  7^"  already  found  to  be  necessary,  we  have 

7J._L  911 —  915 
'4  T^  -16  —  ^16 

or  the  beam  would  need  to  be,  say  10"  x  16". 

Size  of  Strut.  Xow  the  size  of  B  C  must  be  made  sufficient  not 

to  crush  in  the  soft  underside  of  the  beam  at  B.     The  bearing  here 

would  be  across  the  fibres  of  the  beam,  and  we  find  (Table  lY)  that 

the  safe  compressive  stress  of  Georgia  pine  across  the  fibres  ij 

/  — ^  )  ^=  200  pounds.     We  need  therefore  an  area 

25500       ,„Q  .     , 

a  = =128  inches. 

200 

As  the  beam  is  only  10"  wide  the  strut  B  C  will  have  to  measure, 

128 

=z  12+  inches  the  other  way,  or  we  will  say  it  could  be  10"  x  12". 

10  ^  •"  •' 

This  strut  itself  might  be  made  of  softer  wood  than  Georgia  pine,  say 

of  spruce  ;  the  average  compression  on  it  is 

25500       „,„  ,  .     , 

=212  pounds  per  s(iuare  men. 

10.12  ^  ^  ^ 

Xow  spruce  will  stand  a  compression  on  end  (Table  IV)  of 

(  —^  )r=r650  or,  even  if  spruce  is  used,  the  actual  strain  would  be  less 
\J  ^ 

than  one-third  of  the  safe  stress.  At  the  foot  of  the  strut  B  C  yve 
put  an  iron  plate,  to  prevent  the  rod  from  crushing  in  the  wood. 
The  rod  itself  must  bear  on  the  plate  at  least 

Iron  Shoe  to  Z — i— =rr:2,l    square  inches,  or  it  would  crush  the 

Strut.        12OU0  '■ 

iron  —  (120U0  pounds  being  the  safe  resistance  of  wrought-iron  to 

crushing). 

Size  of  Tie-rod.         The  safe  tensional  stress    of  wro  :ght-iron    being 

1  2000  pounds  per  s<iuare  inch  (Table  IV),  we  have  the  necessary 

area  for  tie-rod  A  C  from  Formula  (<>) 

34638  =  a.  12000  or 

a=  — :=  2,886  square  inches. 

12000  '  ^ 

From  a  table  of  areas  we  find  that  we  should  re([uire  a  rod  of 
1  \l"  diameter,  or  say  a  2"  rod. 

The  area  of  a  2"  rod  being  =3,14  square  inches  the  actual  ten- 
sional stress,  per  sipiare  inch  on  the  rod,  will  be  only 

?i^  =  11312  pounds  per  scniare  inch. 


SIZE    OF    ROD.  235 

Size  Of  Washer.  We  must  now  proportion  the  bearing  of  the  wasli- 
er  at  "A"  end  of  tic-rod.  Tlie  amount  of  tlic  crusliing  coming  on 
washer  will  he  whichever  of  the  two  strains  at  A,  (viz.  B  A  and 
A  C)  is  the  lesser,  or  i?  ^  in  our  case,  which  is  32300  pounds.  Wc 
must  therefore  have  area  enough  to  the  washer  not  to  crush  the  end 
of  beam  (or  along  its  fibres),  the  safe  resistance  of  which  we  already 

found  to  be:  T —^  j  =  750  pounds  per  squave  inch;  we  need  there- 
fore 

32300        .„  .     , 

■  =  43  square  inches. 

The  washer  therefore  should  be  about 

Upset  Screw-  "^'^^  ^"^^  *-*^  *^^^  ^'°^  must  have  an  "upset"  screw- 

end,  end ;  that  is,  the  threads  are  raised  above  the  end  of 
rod  all  around,  so  that  the  area  at  the  bottom  of  sinkage,  between 
two  adjoining  threads,  is  still  equal  to  the  full  area  of  rod.  If  the 
end  is  not  "  upset "  the  whole  rod  will  have  to  be  made  enough  larger 
to  allow  for  the  cutting  of  the  screw  at  the  end,  which  would  be  a 
wilful  extravagance. 

Tt  is  unnecessary  to  calculate  the  size  of  nuts,  heads,  threads,  etc., 
as,  if  these  are  made  the  regulation  sizes,  they  are  more  than  amply 
Central  Swivel.  strong.  It  should  be  remarked  here  that  in  all 
trussed  beams,  if  there  is  not  a  central  swivel,  for  tightening  the  rod, 
that  there  should  be  a  nut  at  each  end  of  the  rod ;  and  not  a  head  at 
one  end  and  a  nut  at  the  other.  Otherwise  in  tightening  the  rod 
from  one  side  only  it  is  apt  to  tip  the  strut  or  crush  it  into  the  beam 
on  side  being  tightened.  We  nmst  still  however  calculate  the  verti- 
cal shearing  across  the  beam  at  the  supports,  which  we  know  equals 
the  reaction,  or  20400  pounds  at  each  end.  To  resist  this  we  have 
10"  X  Id"  =  160  square  inches,  less  3"  x  16",  cut  out  to  allow  rod 
end  to  pass,  or  say  112  square  inches  net,  of  Georgia  pine,  across  the 

grain ;  and  as  r -^  j  =  570  pounds  per  square  inch  (see  Table  IV); 

the  safe  vertical  shearing  stress  at  each  support  would  be  (Formula  7) 
112.570:=  63840  pounds  or  more  than  three  times  the 
Bearinsof  actual  strain.      Then,  too,  Ave  should  see  that  the 

Beam,  bearing  of  beam  is  not  crushed.     It  bears  on  each  re- 
action 16  inches,  or  has  a  bearing  area=  10.10  =  160  scjuare  inches. 


236  SAFE   BUILDING. 

( -£j  )  for  Georgia  pine,  across  the  fibres,  Table  IV,  is 

('-£;.  j  =  200,  therefore  the  beam  will  bear  safely  at  each 

end 

160.200=  32000  pounds  or  about  one-half  more  than  the 
reaction.  There  will  be  no  horizontal  shearing,  of  course,  except  in 
that  part  of  beam  under  transverse  strain,  and  this  certainly  cannot 
amount  to  much.     The  beam  is  therefore  amply  safe. 

_  ,,     ^._  Now  let  us  calculate  the  deflection.     The  modulus 

Deflection  , 

of  Beam,  of  elasticity  for  Georgia  pine,  Table  IV  is: 
ez=  1200000  pounds-inch.  The  average  compression  strain  in  A  F 
was  750  pounds  per  square  inch,  therefore  the  amount  of  contraction 
(Formula  88)  i 

x  =  -I^MOi  =  0,19  inches. 
1200000         ' 

Now  A  D  (in  Fig.  14G)  will  be  one-half  of  this,  or  0,095  inches. 

The  amount  of  elongation  in  A  C  will  be,  remembering  that  we 
found  the  average  stress  to  be  only  11312  pounds  per  square  inch, 
and  that  for  wrought-iron  e  =  27000000  (Formula  88) 

11312. 1G3       ^  Ai-oo 

x  = =  0,0o82 

27000000  ' 

The  exact  length  of  yl  C  (Fig.  147  should  be  1G3,41  not  163"). 
Therefore  D  F  (Fig.  140)  will  be 

DF=  163,41 +  0,0682  =  163,4782"    , 

X)£  =  152  — 0,19 
=  151",  81 

Therefore  (Fig.  146) 

2  

^i^=\/ 163,4782  2  — 151,812 
=  60",  655 
Now  B  C  (Fig.  147)  would  be  =  60",  deducting  this  from  the  above 
we  should  have  a  deflection  =  0",  655. 

To  this  we  must  add  the  contraction  of  B  C.  The  strut  will  be 
less  than  60"  long,  say  about  50''.  The  average  compressive  stress 
per  square  inch  we  found  =:  212  pounds.     The  modulus  of  elasticity 

1  111  reality  the  contractiou  of  A  i**  would  be  much  less,  as  the  part  figured  for 
transverse  strain  only  would  very  materially  help  to  resist  the  compression,  one- 
half  of  it  beiDg  in  tension. 


TABLKS    XIX    TO    XXV.  237 

for  spruce,  Table  IV,  is  e  =  850000,  therefore  contraction  in  strut 
(Formula  88) 

d^::^^  =0,0125 

850000 
Adding  tliis  to  the  above  we  should  have  the  total  deflection 
3  =  0,G55-f  0,0125 
=  0,GG75 
This  would  be  the  amount  we  should  have  to  "  camber  "  up  the 
learn,  or  say  J". 

The  safe  deflection  not  to  crack  plastering,  would  be  (Formula  28) 

S  =  L.  0,03 
=  24.0,03 
=  0,72 

So  that  our  trussed  beam  is  amply  stiff. 
Explanation  of  Table  XIX  gives  all  the  necessary  data  in  regard 
?o^xxv.''"'  to  the  use  of  Tables  XX,  XXI,  XXII,  XXIII, 
XXIV  and  XXV.  These  tables  give  all  the  necessary  information 
in  regard  to  all  architectural  sections  which  are  rolled.  Where,  after 
the  name  of  the  company  rolling  the  section  there  are  several  letters, 
it  means  that  practically  the  same  section  is  rolled  by  several  com- 
I-sections  not     panies.     It  should  be  remarked  that  except  in  the 

economical,  ^.j^gg  ^f  ^[^q  simplest  kind  of  beam  work,  it  is  cheaper 
to  frame  up  plate  girders,  or  trusses,  of  angles,  tees,  etc.,  as  there  is 
a  strong  pool  in  the  rolling  of  I-beams  and  channel  sections,  which 
keeps  the  price  of  these  two  sections  unreasonably  high,  in  pro- 
portion to  other  rolled  sections.  Steel  beams  and  sections  are  sold 
as  cheap  as  iron,  (they  are  really  cheaper  to  manufacture),  and 
where  their  uniformity  can  be  relied  on,  should  be  used  in  prefer- 
ence, as  they  are  much  stronger  and  also  a  trifle  stifEer.  As  a  rule, 
however,  the  uniformity  of  steel  in  beams  and  other  rolled  sections 
cannot  be  relied  on. 

One  example  of  an  iron  beam  will  make  the  application  of  the 
Tables  to  transverse  strains  clear,  and  help  to  review  the  subject,  be- 
fore taking  up  the  graphical  method  of  calculating  transverse  strains. 

Example. 

Use  of  Tables  ^  wrought-iron  I-beam  of  25-foot  clear  span,  car- 

XIX  to  XXV.         j.igg  a  uniform  load  of  500  ^^ounds  per  foot  including 
weight  of  beam;  also  a  concentrated  load  o/lOOO  pounds  \Q  feet  from 


238 


SAKE    KL'ILDING. 


the  right  hand  support.    The  beam  is  not  supported  sideways.     What  size 
heam  should  he  used? 

The  total  uniform  load  m  =  500.25  =  12500  pounds  of  which  one- 
half  or  G250  pounds  will  go  to  each  reaction ;  of  the  1000  pounds  load 
ISO 
300 


or  I  will  go  to  the  nearer  support  q  (Formula  15),  therefore 


7  =  6250  +  f.  1000  =  0850 
Similarly  we  should  have  (Formula  14) 

p  =  6250  +  f .  1000  =  G650 
As  a  check  the  sum  of  the  two  loads  should  =  13500,  and  we  have, 
in  effect : 

G850-f  0650=13500 
To  find  the  point  of  greatest  bending  moment  begin  at  q  pass  to  load 
1000,  and  we  will  have  passed  over  ten  feet  of  uniform  load  or  5000 
pounds,  add  to  this  the  1000  pounds  making  6000  pounds,  and  we  still 
are  850  pounds  short  of  the  reaction,  we  pass  on  therefore  towards  j3 
one  foot,  which  leaves  350  pounds  more,  and  pass  on  another  ^'^  of 


l&o 


-l2o-   -"■ 


[~^    5oo]i>s.}>er  ft.'  la^oolhs. 


l6o 


A 


l4o 


(f't>656) 


3oo 


('\'6€>^6) 


Fig.   148. 


afoot  (to  A)  which  very  closely  makes  the  amount.  The  point  of 
o-reatest  bending  moment  therefore  is  at  A,  say  1'  8"  to  the  left  of 
the  weight,  or  140"  from  q:  As  a  check  begin  at  p  and  we  must 
pass  along  160''  or  13'  4"  of  uniform  load  before  reaching  the  point 
i4,  at  500  pounds  a  foot  this  would  make  13^.500.  =  6666  or  close 
enough  to  amount  of  reaction  p  for  all  practical  purposes. 

The  uniform  load  per  inch  will  be^  =41§  pounds. 

Now  the  bending  moment  at  A  will  be,  taking  the  right-hand  side 
(Formula  24) 

??i^=6850.140— 41  §.140.70  — 1000.20 
=  530  667  pounds-inch. 
As  a  check  take  the  left-hand  side  (Formula  23) 
nij,  =  6650.160  —  41f.lG0.80 


USE   OF    TABLKS.  230 

=  530014  j)oun(Is-inch,  or  near  enough  alike  for  all 
practical  purposes. 

Now  the  safe  modulus  of  rupture  for  wrought-iron  (Table  IV)  is 

(-rr  y^=  12000  pounds,  therefore  the  recpiired  moment  of  resistance 

r  from  Formula  (18) 

,._530GG_7_ 
12000 
Looking  at  the  Table  XX  we  find  the  nearest  moment  of  resistance 
to  be  46,8  or  we  should  use  the  12"  — 120  pounds  per  yard  I-beam. 
But  the  beam  is  unsupported  sideways.  The  width  of  top  flange  is 
h  =  5^".  We  now  use  Formula  (78)  to  find  out  how  much  extra 
strength  we  require. 

Reduction  for  In  inserting  value  for  y,  we  use  the  second  column 

uref"^^      ^^      °^  Table  XVI,  as  the  beam  is,  of  course,  of  uniform 
cross-section  throughout,  and  have 
y  =  0,0102. 
In  place  of  to  we  can  insert  the  actual  value  r  of  the  beam,  and  see 
what  proportion  of  it  is  left  to  resist  the  transverse  sti-ength,  after 
the  lateral  flexure  is  attended  to, 


or  r,= 


,  0,0192.252   1  _|_  0,3966 

^  46,8  _, 
'   l,3yGG 

enough.  The  next  size  would  be  the  12-^"  — 125  pounds  per  yard 
beam,  but  as  the  15" — 125  pounds  per  yard  beam  would  cost  no 
more  and  be  much  stronger  we  will  try  that.  Its  width  of  flange  is 
b  =  5"  and  moment  of  resistance  ?-  =  57,93.  Inserting  these  values 
in  (Formula  78)  and  using  r  in  place  of  lo  we  have 

_  57,93  5  7.93 

^' "~  J    I    0,0192"."252  —  1,48 
'  P 

=  39,14 
The  required  moment  of  resistance  was 

r^=:44,2  so  that  this  is  still  short  of  the  mark,  and  we 
should  have  to  use  the  next  section  or  the  15" — 150  pounds  per  yard 
beam.     The  moment  of  resistance  of  this  beam  is  r^69,8  its  width 
of  flange  the  same  as  before,  therefore : 
G9,8        ,„, 
'       1,48  ' 


240  SAFE    CUILDIXG. 

Or  this  beam  would  be  a  trifle  too  strong  even  if  unsupported  fide- 
ways.  We  need  not  bother  with  deflection,  for  the  length  of  beam  is 
only  1§  times  the  span,  and  besides  not  even  f  of  the  actual 
transverse  strength  of  the  beam  is  required  to  resist  the  vertical 
strains,  and,  of  course,  the  deflection  would  be  diminished  accordingly. 

»  .    ..  -r  ,^  The  column  in  Table  XX  headed  "  Transverse 

Safe  Uniform  ,      ,    •  i      -c 

Load.  Value,"  gives  the  safe  uniform  load,  in  pounds,  it 

divided  by  the  span  in  feet,  for  beams  supported  sideways.  Of 
course  the  result  should  correspond  with  Table  XV,  except  that  the 
uniform  load  will  be  expressed  in  pounds  here,  while  it  is  expressed 
in  tons  of  2000  pounds  each  in  that  table.  For  Tables  XXI,  XXII, 
XXIII,  XXTV  and  XXV  the  use  of  the  "  Transverse  Value  "  is 
similar,  and  as  more  fully  explained  in  Table  XIX. 


CHAPTER  VII. 


Gbaphical  Analysis  of  Tkaxsverse  Strains. 


W 


Fig.   149. 


LL  the  (lif- 
erent cal- 
culations 
to  ascertain  the 
amounts  of 
;^J*  bending-mo- 
ments,  the  re- 
quired moments 
of  resistance 
and  inertia,  the 
amounts  of  re- 
actions, vertical 
shearing  on 
beam,  d  e  fl  e  c- 
tions,  etc.,  can 
be  done  graph- 
ically, as  -well 
as  arithmetical- 
ly. In  cases  of 
complicated 
loads,  or  where 
it  is  desired  to 
economize  by 
reducins:    size 


of  flanges,  the  graphical  method  is  to  be  preferred,  but  in  cases  of 
uniform  loads,  or  where  there  are  but  one  or  two  concentrated  loads, 


242  SAl-E    BUILDING. 

the  arithmetical  method  will  probably  save  thne.  As  a  check,  how- 
ever, in  important  calculations,  both  methods  might  be  used  to  advan 
tage. 

Basis  of  Cra-  If  ^^  have  three  concentrated  loads  w,  w„  and  to,, 

phical  Method.  ^^  g^  beam  A  D  (Fig.  149),  as  represented  by  the 
arrows,  we  can  also  represent  the  reactions  p  and  q  by  arrows  in  op- 
posite directions,  and  we  know  that  the  loads  and  reactions  all 
counterbalance  each  other.  The  equilibrium  of  these  forces  will  not 
be  disturbed  if  we  add  at  £■  a  forcc  =  -|-Z/'  pi"oviding  that  at  Fwe 
add  an  equal  force,  in  the  same  line,  but  in  opposite  direction  or  = 

—y- 

We  have  now  at  E  two  forces,  -f  y  and  p.  If  we  draw  at  any  scale 
a  triangle  a  o  x  (or  I)  where  a  o  parallel  and  =p,  and  where  o  x  paral- 
lel and  =  -[-?/,  we  get  a  force x a,  which  would  just  counterbalance 
them,  or  a  x,  which  would  be  their  resultant.  That  is,  a  force  G  E 
thrusting  against  E  with  an  amount  a  x  (or  x,)  and  parallel  a  x  would 
have  the  same  effect  on  E  as  the  two  f orccs  +  ?/ and/).  Continuing 
X, till  it  intersects  the  vertical  neutral  axis  through  load  w at  G,  wa 
obtain  the  resultant  x.  of  the  two  forces  acting  at  G,  namely  x,  and  w 
(see  triangle  6  a  x  or  II).  Similarly  we  get  resultant  x,  at  H,  of  load 
t(j,andx2,  (see  triangle  c  &  x  or  III)  ;  also  resultant  x,  at  I  of  loadti-,, 
and  Xj  (see  triangle  dcxov  IV)  ;  and  finally  resultant  -{-y,^tF  of  re- 
action of  q and Xi  (see  triangle odxovY).  As  this  resultant  is  +  ?/ it 
must,  of  course,  be  resisted  by  a  force  — ?/ that  the  whole  may  remain 
in  equilibrium.  By  comparing  the  triangles  I,  II,  III,  IV  and  V,  we 
see  that  they  might  all  have  been  drawn  in  one  figure  (Fig.  150)  for 
q  -\-p  =  ii\,  -\-  IV,  -{-  w,  therefore : 

do-{-oa=idc-\-cb-\-ba, 
further  both  V      and  IV  contain  J  x  =  x^ 
u        "      V      "      I         "      ox  =  y 
«        «       n     "       I  "       ax  =  x, 

"        "      II     "       III       "       bx  =  X2 
"         "        III    "       IV        "       CX  =  X3 

We  know  further  that  the  respective  Unes  are  parallel  with  each 
other. 

In  Fig.  150  then,  we  have  dc  =  ii\, 
cb  =  u\ 
ba=.w 
ao=-p  and 
od^=q 


roi.E   AXl)    LOAD    LIXK.  243 

The  distance  xy  of  pole  x  from  load  line  da  being  arliitrary,  and 
the  position  of  jiole  x  tlic  same.     Tlie  figure  E  G  H I  F  E  (Fig  149) 
has  many  valuable  qualities.     If  at  any  point  K  of  beam  we  draw  a 
vertical  line  K  L  M,  then  L  M  will  represent  (as 
compared  with  the  other  vertical  lines)  the  pro- 
portionate amount  of  bending  moment   at   A'. 
If  we  measure   L  M  in  parts   of  the   length   of 
A  D  and  measure  xy  (the  distance  of  pole,  Fig. 
150)  in  units  of  the  load  line  da,  then  will  the 
product   of  L M  and  xy  represent   the  actual 
bending  moment  at  A'.     That  is,  if  we  measure 
L  M  in  inches  and  —  (having  laid  out  d  c,  c  h, 
Fig.  150.  etc.,   in  pounds)  —  measure  x  y  in  pounds,  the 

bending  moment  at  A'  will  be^a;  y.  L  M  (in 
pounds-inch.)  Similarly  at  w  the  bending  moment  would  be 

■=x  y.    N  G  (in  pounds-inch.) 
and  at  w;,  it  would  be  ^=x  y.   R  II    "         "         " 
and  at  ty„  it  would  h(i^=x  y.    S  I     "         "         " 
measuring,  in  all  cases,  x  y  m  pounds  and  N  G,  R  H  and  S  I  in 
inches. 

Average  Strain        The  area  of  E  G  II  IF E,  divided  by  the  length 
Fibres.  ^^  span  in  inches  will  give  the  average  strain  for  the 

entire  length  on  extreme  top  or  bottom  fibres  of  beam,  providing  the 
beam  is  of  uniform  cross-section  throughout.  The  area  should  be 
figured  by  measuring  all  horizontal  dimensions  in  inches,  and  all 
vertical  dimensions  in  parts  of  the  longest  vertical  {R  Hin  our  case), 

this  longest  vertical  being  considered  =  (  — ^  )  ^or  top,  or  (— ;- j  for 

bottom  fibres,  or  where  these  are  practically  equal  =  (-^j. 

The  greatest  bending  moment  on  the  beam  will  occur  at  the  point 
where  the  longest  vertical  can  be  drawn  through  the  figure.  From 
this  figure  can  also  be  found  the  shearing  strains  and  deflection  of 
beam,  as  we  shall  see  later. 

Distance  of  If  now  instead  of  selecting  arbitrarily  the  distance 

'*°       X  y  oi  the  pole  from  load  line  d  a  (Fig.  150)  we  had 

made  this  distance  equal  the  safe  modulus  of  rupture  of  the  material, 

or  X  y=(—^  \  —  measuring  x  y  in  pounds  at  same  scale  as  the  load 
line  (Z  a — it  stands  to  reason  that  any  vertical  through  the  Figure 


SAKE    BUILDING. 


E  G  H I F  E  (Fig.  140)  measured  in  inches,  will  represent  the  re- 
quired moment  of   resistance,  for  if   L  M.  x  7/  =  m,  we  know  from 


Fig.    151. 
Formula  (18),  that  ?n  =  ?•.(__  j  and  as  we  made  x  y=  (  —  J,  we 
have,  inserting  values  in  above : 


L3/.(|)=,.(|)or 


LM=. 


SEVKRAL    LOADS.  245 

Having  thus  sliown  the  basis  of  the  graphical  method  of  analyzing 
transverse  strains,  wo  will  now  give  the  actual  nie'Jiod  without  wasting 
further  space  on  proofs. 
Several  Concen-     ^^  there  are  three  loads  tc,u\  and  ?o„  on  a  beam 

trated  Loads.  ^  ^  (Fig.  151)  we  proceed  as  follows  :  at  any  conven- 
ient scale  —  to  be  known  as  the  pounds-scale  —  lay  off  in  pounds, 
dc=u\^;  ahocb  =  w,  and  ba=zw.  Let  A  i3r=Z  measured  in  inches - 
this  scale  being  called  the  inch-scale.  Kow  select  pole  x  at  random, 
Strain  Diagram,  but  at  a  distance    (measured   with    pounds-scale) 

xy=(-^J=:the  safe  modulus  of  rupture  of  the  material.  Draw  x  d, 
xc,  xb  and  x  a.  Now  begin  at  any  point  G  of  reaction  q,  draw  G  F 
parallel  d  x,  till  it  intersects  vertical  w,,  at-F:  then  from  F draw F^ 
paralk-l  c  x  to  vertical  w, :  then  draw  E  D  parallel  6  x  to  vertical 
w ;  and  then  D  C  parallel  a  a;  to  reaction  p.  From  C  draw  C  G,  and 
through  X  draw  x  o  parallel  C  G. 
Reactions.  We  now  have  the  following  results  : 

0  J  =  reaction  q  (measured  with  pounds-scale.) 
ao=      "        p        "  "         "  " 

any  vertical  through  figure  C DE  F  G  C,  (measured  with  inch-scale) 
gives  the  amount  of  rr=  required  moment  of  resistance  in  inches,  at 
point  of  beam  where  vertical  is  measured.  The  longest  vertical 
passes  through  the  point  of  greatest  bending-moment  in  beam,  ^lul- 
tiply  any  vertical  (in  inches)  with  x  y  (in  pounds)  to  obtain  amount 
of  bending-moment  at  point  of  beam  through  which  vertical  passes. 

Moment  of  or  we  should  have :  r=:  y  (92) 

Resistance.  ^     ■' 

Where  r  =  the  required  moment  of  resistance,  in  inches,  at  any 
point  of  beam,  provided  pole  distance  xy=z(JL\. 

Where  v  =:  the  length  (measured  with  inch-scale)  of  the  vertical 
through  upper  figure  C D  EFGC  at  pohit  of  beam  for  which  r  is 
sought. 
And  further : 

Bending-  „j__^,3.„  --ggx 

moment.  ''  ^      ■' 

Where  ni  =z  the  bending  moment  at  any  point  of  beam  in  pounds- 
inch. 

Where  u  =  the  same  value  as  in  Formula  (02) 

Whei-e  x  ?/ =  the  length,  (measured  with  jiound-scale)  of  distance 
of  jiole  X  from  load  line,  in  upper  strain  diagram  x  a  d. 


■246  SAFE   BUILDING. 

If  now  we  draw  horizontal  lines  through  d,  c,  b  and  a ;  and  through 

■0  the  horizontal  line  for  horizontal  axis  ;  and  continue  these  lines  until 

thev  intersect  their  respective  load  verticals  ic\„  il\  and  tv,  the  shaded 

figure  0,  HIJ KLMN  0  0,  will  give  the  vertical  shearing  strain 

along  beam.     Any  vertical  (as  ii,  S)  drawn  through  this  figure  to 

horizontal  axis  and  measured  with  jjounds-scale,  gives  the  amount  of 

vertical  shearing  at  the  point  of  beam  {R)  through  which  vertical  is 

drawn.     Or, 

Vertical  Cross-  

shearing.  ^  —  ^'n  (94) 

Where  .s  =  the  amount  of  vertical  shearing  strain  in  pounds,  at 
zx\y  point  of  beam. 

Where  u,,  ==  the  length  (measured  with  pounds-scale)  of  vertical 
through  figure  O.HIJ  KLM  N  0  0,  dropped  from  point  of  beam  for 
which  strain  s  is  sought. 

We  now  divide  G  C  into  any  number  of  equal  parts  — say  twelve 
in  our  case  —  and  begin  with  a  half  part,  or 

G  to  1  r=r  12  to  C=:  2V   ^  ^^   ^IsO 

lto2  =  2to3  =  3to4  =  4to5,etc.  =  J2-  ^  ^ 
and  make  the  new  lower  load  line  gc  with  inch-scale  so  that 

Deflection^^^    ^  to  I    =  length  of  vertical  1  e 

further  I     to  II  =  length  of  vertical  2/ 

"       II  "  111=     »        "         "       3  A 

"       III "  IV  =     "        «         "       4  i,  etc.  unti. 

"  XI  "  c  =  "  "  "  1 2  k 
Now  select  arbitrarily  a  pole  z  at  any  distance  zj  from  load  line  g  c. 
Now  draw  below  the  beam  where  convenient  (say  I.  Fig.  151) 
beginning  at  .7,,  the  Hne  g,  e,  parallel  g  z  till  it  intersects  the  pro- 
longation of  1  e  (from  above)  at  e, ;  then  draw  e,/,  parallel  I  2 
tiU  it  intersects  vertical  2/at/;  and  similarly  draw /, /i,  parallel 
II  2;  also  //,  2,  parallel  III  z,  etc.,  to  m^  l\  parallel  XI  z  and  finally 
Jr,  c,  parallel  c  z.  The  more  parts  (/,)  we  divide  the  beam  into,  the 
nearer  will  this  line  g,  e,/,  m^  k,  c,  approach  a  curve.  The  real  line 
to  measure  deflections  would  be  a  curve  with  the  above  fines  as  tang- 
ents to  it ;  we  need  not,  however,  bother  to  draw  this  curve  for  prac- 
tical work.  Now  draw  c,  </,  and  parallel  thereto  z  0.  Divide  g,  c, 
at  o„  so  that^:  g,  o„:  c,  o„  =  c  0  :  go,  then  will  o„  be  the  point 
of  greatest  deflection  along  beam.     This  will  be  further  proven  by 

'  Note  that  the  division  of  the  line  g,  Ou  Ci  is  the  reverse  of  the  division  of  the 
line  g  oc. 


CROSS   SHEARING.  247 

the  fact  that  the  greatest  vertical  (in  lower  figure  I)  will  pass  through 


Fig.   152 


o„,  if    the  real  curve  were  drawn.      The  figure  g,e,fJiJ,m,Kc,g, 
will  measure  the    amount  of    deflection  of    beam  at   all   points   of 


248  SAFE   BUILDING. 

beam.  The  deflection  at  any  point  of  beam  being  proportionate  to 
length  of  its  vertical  through  lower  figure  I.  The  amount  of  this  de- 
flection will  be 


Amount  of  De-  /  k 

flecti 
riite  P 
tance 


flection,  Defi-        ^        ■,,    1    r,  i  \~f  ) 


Where  f  -^  j  =  the  safe  modulus  of  rupture,  per  square-inch,  of 


nitePoleDis-        ^-^\-h-zj\f)  (95) 

e.  i. 

AVhere  O  =  the  deflection,  in  inches,  at  any  point  of  beam,  if  pole 
distance  of  upper  strain  diagram  (x  ?/)  :^  (  -^  j. 

Where  r,  =  the  length  of  vertical,  in  inches,  dropped  from  said 
point  through  lower  figure  I  (see  Fig.  151) 

Where  Z,  =  the  length,  in  inches,  of  each  equal  part  1  to  2,  2  to  3, 
3  to  4,  etc.,  into  which  beam  was  divided,  [in  our  case  l^  =  ^^^  /.] 

Where  i  =  the  moment  of  inertia,  of  cross-section  at  said  point,  in 
inches. 

Where  zy=  the  distance  (measured  with  inch-scale)  of  pole  zfrom 
load  line  in  lower  strain  diagram. 

A 
/ 
the  material. 

Where  e  =  the  modulus  of  elasticity,  in  pounds-inch,  of  the 
material. 

If  we  were  to  so  proportion  the  beam  that  the  moment  of  resis- 
tance at  each  point  would  exactly  equal  the  required  moment  of  re- 
sistance as  found  above,  we  should  have :  ^ 

/  k\ 
Deflection  vary-  ,       •  \  — ?•  I 

ins  Cross-  3_Mrf7AZi  (96) 

section.  cl 

V.-.  e 
2 

Where  o,  i\,  zj,  \—p\^  ^•iid  I  same  value  as  in  Formula  (95). 

Where  i;^  length  of  cori-esponding  vertical  in  upper  figure 
C  D  G  E  C,  (to  vertical  v,  of  lower  Fig.  I)  to  be  measured  in  inches. 

"Where  -=  one-half  the  total  depth  of  beam,  in  inches.     Had  we 


not  made  x  y  =  (  -^  j,  we  should  have 


Deflection  Pole  v..l..Z).xy. 

Distance  arbi-     (\^=^ ^ —  ra-7\ 

trary.  "  e.  i.  {y') 

■  This  would  be  the  greatest  possible  deflection.    If  the  beam  were  not  so  pro- 
portioned, but  of  uniform  cross-section  throughout,  the  deflection  would  be  less. 


DEFLECTION.  24? 

Where  3,  v,,  /„  e,  z  J,  and  i  same  value  as  in  Formula  (95). 

Where  X  2/ =1  the  length  of  pole  distance  from  load  line  in  upper 
strain  diagram,  measured  in  pounds. 

The  same  formulse  and  methods  could  be  applied  to  cantilevers, 
but  for  these  the  arithmetical  calculations  are  so  very  simple  that  it 
would  be  taking  unnecessary  trouble. 

A  few  practical  examples  will  make  all  of  the  foregoing  more  clear. 

Example  I. 

Single  concen-       ^  Georgia  pine  girder  A  B  of  20-foot  span  carries 
trated  Load,  a  load  w,  of  2000  pounds  5'  0"  fro7n  right  reaction  D. 
Wliat  size  should  the  girder  bef 

We  draw  (Figure  152)  A  B  =  2i0"  at  inch-scale,  and  locate  w,  at 
60"  to  the  left  of  B.  Now  draw  a  vertical  line  6  a  =  2000  pounds  at 
pounds-scale.       Select  point    x  anywhere,  but   distant  x  2^  =  1200 

pounds.  (1200  pounds  being  =  ^—^  or  the  safe  modulus  of  rup- 
ture, per  square-inch,  of  Georgia  pine).  Draw  x  b  and  x  a.  Draw 
verticals  through  A,  to,  and  B.  On  vertical  A  begin  at  any  point  C, 
draw  C  E  parallel  x  a,  till  it  intersects  verticals  w,atE;  then  draw 
E  G  till  it  intersects  vertical  B  at  G.  Draw  G  C  and  o  x  parallel  to 
G  C.  We  scale  o  b,  it  scales  1500  pounds,  so  this,  is  the  reaction  at 
B.  We  scale  a  o,  it  scales  500  pounds  and  this  is  the  reaction  at  A. 
The  longest  vertical  through  C  E  G  is  vertical  w„  therefore  greatest 
bending-moment  is  at  w;,  which  we  know  is  the  case.  We  scale  E  D 
at  inch-scale,  it  scales  75  inches,  therefore  the  (greatest)  required 
moment  of  resistance  will  be  at  w^  and  will  be  Formula  (92). 
r=75. 
From  Table  I,  section  No.  2,  we  know  for  rectangular  beams, 

r  =  -^,  therefore ; 
(J 

h.d^      _. 
—  =  .o,or 

b.(P  =  A^O. 

We  will  suppose  the  girder  is  not  braced  sideways,  and  needs  to  be 

pretty  broad  ;  let  us  try  b  =  5",  we  have  then : 

5.  (/"  =  450  or 

,„      450       „.        -. 
d'= —  =  90  and 
5 

d=9,  5"  or  the  girder 
would  have  to  be  5"  x-  9^"  or  say  5"  x  10".     The  bending-moment  at 


250  SAFE    BUILDING. 

w,   is,   of   course,    Formula    (93)  =  E  D.  x  y  =  75.1200  =  90000 
(pounds-iuch). 

Had  we  calculated  arithmetically,  we  should  have  had,  Formulae 
(14)  and  (15)  : 

reaction^  =—.  2000  =  500  pounds. 
240 

"        5=—.  2000  =  1500  pounds. 
240  ' 

Bending  moment  at  w,  would  be  (right  side)  Formulae  (23)  and 
(24).      ?C  =  1500.G0  —  0.2000  =  90000  (pounds-inch)  or  check 
(left)  side  m^,  =  500.180  —  0.2000  =  90000  (pounds-inch.)      There- 
fore required  moment  of  resistance,  Formula  (18) 
_  90000  _^- 
''        1200  " 
or  same  result  as  graphically. 

By  drawing  the  horizontals  from  6  between  verticals  B  and  w^ ; 
from  a  between  verticals  A  and  w, ;  and  from  o  between  verticals  A 
and  B  we  get  the  etched  figure  for  measuring  vertical  shearing 
strains.  We  see  at  a  glance  that  the  shearing  to  the  right  of  load  is 
equal  to  the  right  reaction,  and  is  constant  at  all  points  of  right  side 
of  beam ;  while  on  the  left  side  of  load  it  is  equal  to  the  left  reaction, 
and  is  constant  at  all  points  of  the  left  side  of  beam.  And  this  we 
know  is  the  case.  We  need  not  bother  with  shearing,  however,  for 
we  can  readily  see  there  is  no  danger.  For  even  immediately  to  the 
right  of  the  load,  the  weakest  point  in  our  case,  we  know  that  one- 
half  of  the  fibres  of  cross-section  are  not  strained  at  all,  or  we  should 
have  one-half  of   area  or  ^=25    square-inches   to  resist   1500 

pounds  of  shearing,  or  i|^  =  60  pounds  per  square-inch,  while  the 

safe  resistance,  per  square-inch,  of  Georgia  pine  to  shearing  across 

the  grain  is  (Table  IV)    (iL^  —  bld  pounds. 

There  is,  however,  some  danger  of  excessive  deflection ;  we  draw, 
therefore,  the  figure  c,/,  (/,  by  dividing  the  beam  into  ten  equal  parts, 
beginning  and  ending  with  half  parts  at  the  reaction,  (each  whole 

part  being  24"  long,  or  Z,  =^=  24") 

We  draw  the  verticals  through  these  parts  and  get  their  lengths 
through  figure  C  E  G.  These  lengths  we  carry  down  in  their  proper 
succession  on  the  load  line  r/j  c  of  the  lower  strain  diagram,  begin- 


SINGLE    LOAD. 


251 


ning  at  the  top  with  the  right  vertical  1,  putting  immediately  under 
this  the  length  of  second  vertical  2,  then  3  and  so  on  till  ^c  =  sum 
of  lengths  of  all  ten  verticals  through  C  E  G.  We  now  select  z  at 
random  (in  our  case  120  inches  from  load  line  or  2y  =  120").  "We 
now  draw  lines  from  zio  g  I,  II,  III,  etc.,  to  c.  Construct  figure 
Qifi  '^i^Y  beginning  at  ^,  drawing  Une  parallel  to  z^  until  it  intersects 


Fig.   153. 


prolongation  of  first  vertical  1 ;  then  line  parallel  tozl  till  it  intersects 
prolongation  of  second  vertical  2,  etc.  We  now  draw  z  o  parallel 
c,  ^,.  We  scale  g  o  and  find  it  scales  222",  also  c  o  which  scales  162"; 
we  divide  c,  ^,  at/,  so  that 

cj:fg,  =  222:  1G2. 


252  SAFE    BUILDING. 

Carrying  vertical//,  through  figure  we  find  it  scales  (y,)  =  117"  con- 
tinuingy/,  up  to  beam  it  gives  us  point  Fas  the  point  of  greatest  de- 
flection, we  find  A  F  scales  138".     Had  we  used  Formula  (43)  we 

/  240"  — 60= 
should  have  located  P  at  a  distance  from  A  or  A  F=  -*  / 5 

=  134,  17".     So  that  we  have  a  sufficiently  accurate  result. 

For  the  amount  of  deflection  at  i^we  use  Formula  (95)  ;  we  iinovv 

7,     ^3  K      1  A3 

that  (Table  I,  Section  No.  2)  i  =  ^  =  ^^=  4.17,  further  for 
Georgia  pine  f-^j  =  1200  pounds. 

e  =  1200000  (inch-pounds.) 
/,  =  24" 
v,=ff,  =  117" 
2/=  120",  therefore : 
o  _  11 7.  24. 120.  1200 
1200000.     417 
=  0,808'' 
Had  we  calculated  the  deflection  by  Formula  (41)  we  should  have 
had: 
remembering  that  ?rt  =  180"  and  n  =  60"  and  i  +  n  =  240-l-C0  = 

300"  

o  _  2000. 180.  GO.  300  /I8O.  300 

~~  9.240.1200000.417'    "Y'        ^ 
=  0,803" 
Which  proves  the  accuracy  of  the  graphical  method. 

For  a  beam  of  20  feet  span  the  deflection  not  to  crack  plastering 
should  not  exceed.  Formula  (28). 
8  =  20.0,03  =  0,6" 
Therefore,  if  our  beam  supports  a  plastered  ceiling,  it  must  be  re- 
designed to  be  stiffer.  Either  made  deeper,  in  which  case  it  can  be 
thinner,  if  braced  sideways,  or  it  can  be  thickened  sufficiently  to  re- 
duce the  deflection,  see  Formula  (31). 

Example   II. 
Single  centre  A  hemlock  girder  A  B  (Fig.  153)  of  IQ-foot  span, 

load,  carries, a  centre  load  w  of  1000  pounds.      What  size 
should  the  girder  be  ? 

We  make  A  .2  =  192"  at  inch  scale;  locate  w  at  its  centre  F; 
m.akeftaat  any  scale  —  (pounds-scale)  —  equal  1000  pounds.  Se- 
lect pole  X  distant,  a;?/ =750  pounds,  from  load   line  b  a,  (as  750 


CKXTIIE    LOAD. 


253 


pound3  =  (--;)  the  safe  modulus  of  rupture  per  square  inch  of  hem- 
lock). Draw  xb  and  x  a.  Uugin  at  G,  draw  G  E  parallel  6  x  to 
vertical  through  load,  and  then  draw  E  C  parallel  a  x.  Draw  C  G 
and  then  x  o  parallel  C  G,  we  find  that  o  bisects  6  a  or  a  o=  o  &  =  500 
pounds.  Each  reaction  is  therefore  one-half  of  the  load;  this  we  know 
is  the  case.  Greatest  Une  through  C  G  E  vie  find  is  at  D  E,  so  that 
greatest  bending-moment  is  at  load;  this  we  know  is  the  case.  DE 
scales  64"  at  inch-scale,  therefore  the  recjuired  moment  of  resistance 
for  the  beam  is,  Formula  (92)  : 

r  —  Gi. 
and  the  greatest  bending-moment  at  load,  Formula  (93)  : 
mw  =  G4.  2;?/==G4.750 
=  48000 
Had  we   calculated   arithmetically  we   should  have  obtained  the 
same  results,  for  Formula  (22) 

1000.192        .„„„„ 

Tn„  = =  48000 

4 

and  Formula  (18) : 

48000       ^, 

•  ?•  =  • =  04 

750 

Now  from  Table  I,  Section  No.  2,  we  know  that  for  rectangular 

sections : 

b.(P 

r= or 

6 

&.d2=  64.6  =  384. 
If  we  assume  the  beam  as  4"  thick,  we  have  then  : 
4.^2  =  384. 

^2__384_gg  ^j, 

4 
d=  J96^=9,8"  or  we  will  make  the  beam  4"  x  10". 
We  draw  the  figure  0,  H  J  K  N  0  for  shearing  and  find  it  is  con- 
stant throughout  the  whole  length  of  beam  and  equal  to  length  0,  H 
or  N  0  measured  at  pounds  scale,  or  500  pounds.     This  is  so  small 
we  need  not  bother  with  it. 

To  obtain  the  deflection  diagram  we  divide  G  C  into  eight  equal 

parts,  each  part  1,  =  ^  —  24"  and  begin   at  each  end  with  half 

parts,  drawing  the  eight  verticals  through  C  E  G. 

We  lay  off  their  exact  lengths  in  proper  succession  on  the  lower 
load  line  g  c,  beginning  at  the  top  with  the  right  vertical.     Select 


254 


SAFE    BUILDING. 


pole  z  at  random,  in  this  case  distant  from  load  line  2y=:180".  We 
now  draw  the  figure  c,  ^,/,  and  find  greatest  deflection  is  at  its  cen- 
tre//,; for  z  0  parallel  c,  </,  bisects  gc.     We  scale  //  at  inch  scale 


Fig.    154. 


=  44",  therefore  greatest  deflection  of  beam  at  centre,  Formula  (95), 
remembering  that   (Table  I,  Section  No.   2)  t=-L^  =  333  and 


TWO   I.OADS.  255 


(Table  IV)  c=  800000 

3^44.24.180.750^ 

800000.  33:i  ' 

Had  we  calculated  the  deflection  arithmetically  from  Formula  (40) 
we  should  have  had  : 

3^1    1000.1 02«_^ 
48  800000.333         ' 
or  practically  the  same  result. 

If  the  beam  supported  plastered  work  the  deflection  should  not  ex- 
ceed, Formula  (28) 

8  r=lG.  0,03  =  0,48" 
Still,  unless  we  were  very  particular,  the  beam  could  be  passed  as 
practically  stiff  enough. 

Example  III. 

Two  concen-  A  loliile  jiine  beam  A  B  Fig.  154,  of  12-foot  span 

trated  loads,  carries  two  loads,  one  w,  =  800  pounds,  four  feet  from 
left  support,  the  other  w'„  =  1200  pounds,  two  feet  from  right  support. 
What  size  should  the  beam  he  ? 

Make  ^  JS  at  inch  scale  =  144  inches,  locate  w,  so  that  A  w,  =  48", 
and  w,,  so  that  B  w„  =  24".  At  any  (pounds  scale)  make  b  c=:  1200 
pounds  and  c  a^^  800  pounds.  Select  pole  x  distant  from  b  a ; 
a:?/  =  900  jwunds,  the  safe  modulus  of  rupture  per  square  inch  of 
white  pine ;  draw  xb,  xc  and  x a.  Construct  C DE  G  parallel  to 
these  lines.  Draw  C  G,  and  parallel  to  same  xo,  then  will  a  o  =  733 
pounds  be  reaction  at  A,  and  o&:=1267  pounds  be  reaction  at  B. 
We  scale  vertical  D  N  at  iv,  =  3d"  and  T E  at  w„=35",  therefore 
greatest  bcnding-moment  is  at  to,  and  Formula  (93) 
m«.,==39.  900  =  35100 

Further,  the  required  moment  of  resistance  at  t(7,  Formula  (92) 
will  be : 

r  =  DN=3d. 

Now  from  Table  I,  Section  No.  2, 

b.d^ 

r= — — ,  or 

6 

&.rf2  =  6.39  =  234 

Now  if  6  =  3"  we  should  have 

cZ2  =  !M=78and 
3 

d  =i/78  =  say  9",  or  the  beam  would  need  to  be  3"  x  9". 


256  SAFE    BUILDING. 

We  should  have  obtained  praetically  the  same  results  arithmeti- 
cally, for  :  Formulae  (IG)  and  (17)  : 

,       800.96   .1200.24 
Reaction  atA=   ^^^    -p  — jj^ — =  ^33 

„       800.48  ,    1200.120       ,„^^ 
Reaction  at^=     ,  . .    "r  — Y4I —  ^^  ^^" 

check:  A  -\-  B  =  w,  ^  w,,=  800  +  1200  =  2000  pounds  and 
733  +  1267  =  2000  pounds. 

Beginning  at  B  we  have  to  pass  over  load  iv,,  (1200  pounds)  and 
on  to  w„  before  passing  amount  of  reaction  B  (12G7  pounds)  there- 
fore greatest  bending-moment  at  ii\.  We  know  from  Formula  (24) 
it  would  be : 

m„,=  1267.96  —  72.1200  =  35232 
and  check  from  Formula  (23) 

m^,  =  733.48  —  0.800  =  35184 
being  near  enough  for  practical  purooses.     From  Formula  (18)  we 
should  have  had : 

35184^3 
900  ' 

We  now  draw  the  shearing  diagram  0,  H  IJ  K  L  M  0,  as  shown 
in  Figure  154,  and  find  the  amount  of  shearing 
from  A  to  w,=  0,  H=    733  pounds, 
from  w,  tow,,  =  J  S  =      67  pounds, 
from  w„  to  B=MO=V2G7  pounds. 

We  can  overlook  it,  for  even  at  the  weakest  point  of  beam  for  re- 

3  9 
sisting  cross-shearing  we  have  half  the  area,  or-^  =  13|^  square 

inches. 

White  pine  will  safely  resist  250  pounds  per  square  inch  in  cross- 
shearing  (Table  IV)  or  the  beam  would  resist. 

13|.  250  =  3375  pounds  at  its  weakest  point  for  cross-shearing, 
(viz.:  at  10,)  and  twice  as  much  at  the  reactions. 

To  find  the  deflection  we  divide  G  C  into  eight  equal  parts,  begin- 
ning with  half  parts  (or  7,=—=  18")   and    draw    the    verticals 

through  C  D  E  G.  We  now  make  the  lower  load  line  g  c  equal  the 
sum  of  these  verticals,  beginning  at  the  top  with  the  right  vertical. 

Select  z  distant  from  g  c  (the  load  line)  zj=.  108".  Draw  zg,zc, 
etc.,  and  construct  g,  c^f^  as  before. 

We  draw  z  0  parallel  r,  </,.     Now  g  0  measures  116  inches  and  0  c 


TWO   LOADS.  257 

108  inches,  therefore  divide  c,  g^  at/ so  that : 
cJ:fg.  =  lU:  108 
Carrying  the  vertical//,  up  to  point  Foi  beam,  we  find  the  point 
of  greatest  deflection  F,  where 

B  F=  G9i"  and  A  F=  74^" 
We  find//,  scales  42",  remembering  that  (Table  I,  Section  No.  2) 

{=±±  =z  182,  and  that  for  white  pine  Table  IV  e=  850000  pounds 

we  have  Formula  (95) : 

3^42.18.108.900^ 

"          850000  .182  ' 

Had  we  attempted  to  get  this  result  arithmetically  by  inserting  the 
values  in  Formula  (41)  (and  remembering  that  n  is  always  the  nearer 
support,  or  in  our  case  respectively  48"  and  24",  while  m  respectively 
96"  and  120")  we  should  realize  the  advantage  of  the  graphical 
method,  for :  

_800.96.48.(144  +  48).V/"'^-<^^^3+^^>  +  1200.96.48.(144  +  24).V/^gg^^gi> 
*'"■  9.144.850000.182 

If  we  figure  out  the  above  tedious  formula  we  should  have 
8  =  0,422" 
or  practically  the  same  result  as  we  obtained  graphically. 

The  safe  deflection,  were  the  beam  to  carry  plastering,  should  not 
exceed  Formula  (28) 

8=12.0,03=0,36" 
Our  beam  is  therefore  not  nearly  stiff  enough,  and  we  must  make  it 
thicker  ;  or  else  if  we  wish  to  save  material,  we  will  make  it  thinner, 
but  deeper ;  and  then  brace  it  sideways,  see  Formula  (31). 

Example  IV. 

Five  Concentra-     A  spruce  girder  A  B  of  18/oo<  span  carries  five 
ted  Loads,  i^adg^  ^g  sliown  in  Figure  155.     What  size  should  the 
girder  be  f 

We  draw  AB  =  2W  (inch  scale) ;   further 

6a=:2700  pounds  (sum  of  loads  at  pounds  scale)  ;  make 

bh  =  Wy  =  540  pounds, 

ii  g  =  w,v  =  1 80  pound?, 

ed  =  w,,,  =  360  pounds, 

dc  =  to„  =720  pounds,  and 

c  a  =  iOi  =  900  pounds. 


258 


SAFE   BUIT-DING. 


At 
C 


:4& 


?»^-°^ 


ooooo«o«<> 


Fig.    155. 


FIVE    CONXENTUATKI)    LOADS.  259 

Select  X  distant  x  y^  1000  pounds  from  h  a,  (as  1000  =^(  -^)  for 

spruce,  see  Table  IV).      Draw  x/),  x//,  are,  etc.,  and  figure  C  D  G. 
Draw  xo  parallel  C  G;  it  divides  load  line  as  follows: 
a  0=- 1580  pounds  or  reaction  at  A. 
o&  =  1120  pounds  or  reaction  at  B. 
We  find  longest  vertical  through  C  D  G,  is  at  load  w,„  therefore 
greatest  bending-moment  on  beam  at  w,^ ;  now  D  E  scales  70^",  there- 
fore Formula  (93)  : 

m,y„  =  70i.  1000^:70500 
and  Formula  (92) 

?-=70,  5 
From  Table  I,  Section  No.  2 

b  il^ 
r  =  -^-  =  70,  5  and  ii  b  =  5,  we  have 
6 

5.  c?2  =  6.  70,  5  or 

rf2  =  84,Gand 

d  =i/84,  6  =9,2"  or  say  10"  which   is  the  nearest  size 

larger  than  9,2",  and   of  course  Avooden  beams  are  never  ordered  to 
fractions  of  inches. 

Had  we  worked  arithmetically  we  should  have  had  practically  the 
same  rer-ults. 

From  Formulae  (IG)  and  (17)  we  should  have  had: 
reaction  at  A  =  1580  pounds, 
reaction  at  jB  =  1 1 20  pounds. 

From  rule  for  finding  greatest  bending-moment  we  should   have 

located  it  at  w„  and  then  had  Formula  (23) 

TOv^,,  =  1580,  72  —  48.900  =  70560 

and  from  Formula  (18) 

705G0        r,n   nr 

r= =  70,  5b 

1000 

We  now  draw  the  shearing  diagram  0,  II I  J  K  L  M  N  P  0  and 
find  as  follows  : 

Cross-shearing  A  to  w,  =11  0  =  1580  pounds. 
Cross-shearing  ^o^  to  20„  =  J  K  =  680  pounds. 
Cross-shearing  w„  toiv^,^=KL=  40  pounds. 
Cross-shearing  2t',„  to tL\y  =  M  E=  400  pounds. 
Cross-shearing  w,v to  Wy  =N  S  =  580  pounds. 
Cross-shearing  tOy  toB    =  P  0  =1120  pounds. 

We  need  not  bother  with  it,  therefore.     For  deflection  we  now^ 


2G0  SAFE   BUILDIXG. 

divide  C  (? again  into  eight  equal  parts,  (or  l^  =  ^^--=27")  beginning 

8 

■with  half  parts  at  C  and  G.     We  now  make  lower  load  line  g  c  = 

the  sum  of  the  eight  verticals,  putting  the  right  vertical  at  the  top 

from  g  down.     We  select  pole  z  at  a  distance  zj  =  120"  from  g  c  and 

draw  zg,  zc,  etc.    We  construct  figure  g,f,  c,  and  draw  zo  parallel  to 

'^i  9i-     ^^6  now  divi  le  c,  (/,  at/,  so  that 

g^f:  fc^r=iCO'.  0 g,  carrying //  up  to  beam,  we  have  the  point 

F,  distant  102"  from  B,  and  114"  from  A,  which  is  the  point  of 

greatest  deflection.     We  find  that//,  scales  102",  remembering  that 

e  =  850000  for  spruce  (Table  IV),  and  that  iz=±^  =417  (See 

Table  I,  Section  No.  2)  we  have,  Formula  (95). 

g^  102.  27. 120.  1000^ 
850000.417 

This  would  be  too  much  for  plastering,  for  if  the  girder  supported 
plastering,  the  deflection  should  not  exceed  Formula  (28) 
S  =  18.  0,  03  =  0,54" 

We  must  therefore  deepen  the  beam  very  materially. 
We  use  Formula  (31), 
_    1 

In  our  case  it  would  be 

x=     ^  „  =  ^— =  0,0002 
5.103       5000 

Supposing  we  were  to  make  the  beam  4"  x  1 2",  then  we  should 

have 

a;  =  -1^  =  0,  000  144 
4.123         ' 

The  deflection  of  the  latter,  then,  would  be 
§:  0,93  =  0,000144:  0,  0002  or 

S  =  0>93.  0.000144^Qg^,, g^jjj  ^^^  ^^^^^  deflection. 
"  0,0002 

Were  we  to  make  the  beam  3"  x  14",  we  should  have  : 

x=— 1^=0,0001215 
3.143 

The  corresponding  deflection  for  this  beam  would  be  : 
S:  0,93  =  0,0001215:  0,0002  or 


g  _  0,93.  0,0001215  _,, 
°  ~  0,0002  ' 


FIVE    CONCENTHATKD    LOADS.  261 

or  just  about  what  would  be  required  iu  the  way  of  stiffness. 


©<g) 


A^ 


1- 


®  ^  ©  ©  ©  ^ 


T>»r  ^•♦V* 


Fig.    156. 

Had  we  used  Formula  (95)  we  should  have  had,  remembering  that 
now 

12 


8  = 


102.27.120.1000 


=  0,5G8" 


850000.686 
showing  that  we  have  made  no  mistake  in  applying  Formula  (31). 


•262  SAFE    BUILDING. 

If  we  have  any  doubts  as  to  whether  a  3"  x  14"  stick  is  as  strong  as 
a  5"  X  10"  we  use  Formula  (30)  and  have  for  tne  former 

a;  =  3. 142  — 588 
while  for  the  latter 

a;  =  5. 102  =  500,  so  that  the  3"  x  14"  stick  is  actually 
much  stronger,  as  well  as  much  stiffer  than  the  5"  x  10".  It  is,  how- 
ever, a  very  thin  beam,  and  would  be  apt  to  warp  or  twist,  unless 
braced  sideways  about  every  five  feet  of  its  length. 

To  attempt  to  get  the  deflection  of  the  girder  arithmetically  would 
be  a  very  tedious  operation.  It  could  be  done,  however,  by  inserting 
in  Formula  (41)  the  different  values  for  n  and  ?«,  remembering  every 
time  to  make  n  the  distance  from  each  weight  to  the  nearer  support 
to  respective  weight,  and  m  the  distance  from  same  weight  to  the 
further  support. 

Example  V. 

Uniform  Load.  A.  wr^aghi-iron  beam  of  25-foot  span  (Figure  156) 
carries  a  uniform  load  of  800  pounds  per  running  foot  of  beam,  in- 
cluding weight  of  beam.  The  beam  is  tliorougldy  braced  sideways. 
What  beam  should  be  used  f 

We  draw  A  B  =  300"  at  inch  scale,  and  then  divide  our  uniform 
load  into  a  number  of  equal  sections,  say  eight,  each 

l,  =  ^  =  d7i"  long. 

The  total  load  on  beam  is 

u  =  25. 800  =  20000  pounds. 
Each  section  therefore  carries : 

U  20000  T-nn  1 

_  = =  2o00  pounds. 

8  8  ^ 

We  place  our  arrows  iv„  w,„  etc.,  at  the  centre  of  each  section, 
which  will  bring  the  end  ones  at  ^  distant  from  each  support,  so  that 

these  same  verticals  will  answer  when  obtaining  deflection  figure. 

We  now  make  i  a=:  20000  pounds  at  pound  scale,  and  divide  it  into 
eight  equal  parts,  each  equal  w^  =  il\,^=io,„  etc.,=  2500  pounds.    We 

make  XT/ =  12000  pounds,  which  is  the r -^  j  for    wrought-iron,    see 

Table  IV.  We  draw  xb,xa,  etc.,  and  construct  figure  C  E  G, 
which  will  approach  a  parabola  in  outline.  The  more  parts  we  take 
the  nearer  will  it  be  to  a  parabola. 

We  draw  x  0  parallel  C  G  and  find  it  bisects  b  a,  or  each  reaction 


UNIFORM   LOAD.  263 

is  one-half  the  load  or=  10000  pounds.  This  we  know  is  the  case. 
The  longest  vertical  will,  of  course,  be  at  the  centre  Z)  of  C  G,  or 
greatest  bending-moment  will  be  at  the  centre,  this  we  know  is  the 
case.  D  E  scales  (inch  scale)  62^"  which  will  be  the  required  r  or 
moment  of  resistance  (Formula  92).  The  bending-moment  at  the 
centre  will  be.  Formula  (93). 

m=G2^.  12000  =  750000 
Had  we  used  Formula  (21)  we  should  have  had 

m  = -^ ==  750000  or  same  result,  and  from  For- 

8 

mula  (18)  for 

r=i  ■ — - — =62i  also  the  same  as  before.      From 
12000  ^ 

Table  XX  we  find  the  nearest  r  to  our  required  r  (62,5)  is  69,8 

which  calls  for  a  15"  — 150  pounds  beam;   as  the  beam  is  braced 

sideways  this  will  do,  if  sufficiently  stiff. 

In  regard  to  shearing,  we  draw  the  figure  O^H  I J  K  L  M  N  P 
R  S  0  and  find  shearing  on  bolh  sides  of  beam  similar,  increasing 
gradually  from  the  centre  to  ends.^ 

It  would  be : 
Cross  shearing  from  A  to  ?<?,  =  0,  H  =  10000  pounds. 
Cross  shearing  from  ?«,  to  u\,  =  T  I  =  7500  pounds. 
Cross  shearing  from  w,^  to  ?t',„  =  V  J  ==  5000  pounds. 
Cross  shearing  from  ?«„,  to  tL\y  =  L  K  =  2500  pounds. 
Cross  shearing  from  w^y  to  Wy    =  0       =  0  pounds. 

Cross  shearing  from  Wy  to  u\r,  =  M  N  =  2500  pounds. 
Cross  sheering  from  zi),.,  to  itVu  =  P  P,  ^=  5000  pounds. 
Cross  shearing  from  Ji^vn  to  Wym  =  R  R,=  7500  pounds. 
Cross  shearing  from  Wy^,^  to  B     =  S  0    =  10000  pounds. 

The  area  of  web  of  a  15"  — 150  pounds  beam  (Table  XX)  is 
7,59  square  inches;  the  safe  resistance  of  wrought-iron  to  cross- 
shearing  per  square  inch  being  (il\=  8000   pounds,   we   need    not 

worry  any  further  on  that  score. 

To  find  the  deflection  we  now  make  the  lower  load  line  g  c  equal  to 
the  sum  of  the  lengths  of  verticals  ?rv„„  Wy,,)  Wvu  etc.,  through  parabola 
C  E  G,  beginning  at  top  g  with  length  of  right  vertical  Wviu*  We  select 
z  at  random,  scale  zj-=^  246"  (inch  scale),  draw  z  g,zc,  etc.,  and  figure 

1  Had  we  taken  more  parts,  the  steps  in  shearing  figure  would  become  smaller 
and  smaller  till  they  would  finally  assume  the  straight  line  II S,  which  is  the 
real  outline  of  shearing  figure. 


264  SAFE    BUILDING. 

<^ifx  9i  •  We  now  draw  zo  parallel  c,  g,  and  find  it  bisects  g  c,  or 
greatest  deflection  will  be  at  centre  of  beam,  which  we  know  is  the 
case.  We  scale //  =  62"  (inch  scale);  find  from  Table  XX  for 
our  15"  — 150  pounds  beam  i=523, 5  and  from  Table  IV  for 
wrought-iron  c  =  27000000,  therefore,  Formula  (95)  : 

o  _  62.37,5.246.12000  _  ^  ^gg„ 

°~~     27000000.523,5  ' 

Had  we  figured  arithmeticallj',  Formula  (39),  we  should  have  had 

cv  _       5.  20000.  3003       _  ^  ^^^,, 

^~  384.27000000.523,5""    ' 
or  practically  the  same  result. 

The  safe  deflection  for  plastering  should  not  exceed  (28) 

8  =  25.0,03  =  0,75" 
80  that  we  are  perfectly  safe,  providing  our  beam  is  well  braced 
sideways. 

Example  VI. 
Uniform  and  J   wrouqht-iron  beam,  braced  sideways,  of  50-foot 

Concentrated  -r,-  ,-^  •  y  ;         ?        ^    nr^n 

Load,  span,  Figure  lo7,  carries  a  unijorm  load  oj  200 
pounds  per  foot,  including  weigM  of  beam.  It  carries  also  a  concen- 
trated load  w^  =  10000  pounds  ten  feet  from  the  right-hand  support. 
What  beam  should  be  used  f 

We  draw  beam  A  5  =  360"  at  inch  scale,  we  divide  uniform  load 
into,  say,  six  equal  parts,  each  5  feet  long,  or  /,=:60".  The  total 
uniform  load  will  be  w  =  30.200  =  6000  pounds,  therefore  each  part 

u  —  !^^lr2,=  1000  pounds.  We  draw  arrows  at  the  centre  of  each 
6~~    6  ^ 

uniform  part,  so  that  the  end  arrows  will  be  one-half  part  from  sup- 
ports. These  arrows  will  therefore  answer  for  our  verticals,  when 
drawing  deflection  figure. 

At  120"  from  right  hand  support  we  locate  the  load  w,  =  10000 
pounds. 

We  now  make  load  line  6  a  :=  16000  pounds  the  total  load  and 
divide  it,  so  that 

6Z  =  2tVn=    1000  pounds. 

I  h  =  Wy,  =    1000  pounds. 

hf=  w,    =10000  pounds. 

fe  =  Wy   =  1000  pounds. 

ed  =  ii\y  =  1000  pounds. 

dc  =  w,„=   1000  pounds. 

ca  =  w,,   =  1000  pounds. 


UNIFORM   AND    COXCENTRATED   LOADS. 


265 


®  ©  ©  a®® 


:::::! j  i  ::?  i  : 


Fig.    157. 


266  SAFE   BUILDING. 

Select  pole  x  distant  from  load  line  at  random  (for  tlie  sake  of  il- 
lustration, though  it  would  be  better  to  make  a:7/  =  (- j  =  12000 

pounds.)  We  find  a;y  scales  6500  pounds.  We  now  draw  a;  5,  zZ,  a;  A,  x^ 

etc.     And  construct  figure  C  E  G.     Draw  xo  parallel  C  G  and  we 

find  a  0  (or  reaction  A)  scales  =  6333  pounds,  and  o  b  (or  reaction  B) 

scales  :=  9667  pounds. 

The  longest  vertical  is  DE=  161"  (inch  scale)  therefore  greatest 
bending-moment  is  at  w,  and  from  Formula  (93) 
m^,  =  161.6500  =  1046500 

For  the  required  moment  of  resistance  we  have  from  Formula  (18) 
1046500       Q„„ 
12000 

The  cheapest  or  most  economical  nearest  section  we  find  —  to  this 
required  r  (87,2)  is  the  20"  —  200  pounds  beam  of  which  the  moment 
of  resistance  is  r=  123,8. 

Had  we  combined  the  formulae  for  uniform  and  concentrated  loads 
and  worked  out  the  problem  arithmetically  it  would  have  been 
tedious,  but  we  should  have  had  similar  results. 

We  can  safely  overlook  shearing,  but  note  that  the  real  shearing 
figure  would  not  be  the  shaded  figure,  but  dotted  figure  0,  H UK  0 

For  finding  the  deflection  we  now  draw  lower  load  line  g  c  =  the 
sum  of  the  verticals  through  C  E  G,  beginning  at  top  with  length  of 
Wv,,,  then  Wy„  w^,  w,^,  w,,,,  and  w,,  in  their  order.  We  take  no  notice 
of  vertical  ii\  as  it  does  not  fall  in  one  of  the  even  divisions  oi  C  G 
or  AB  into  lengths  I,.  We  select  pole  z  distant  zy  =  288"  from  load 
line,  draw  zg,  zc,  etc.,  and  then  figure  c,f,g^.  We  now  draw  2  o 
parallel  c, g,  it  divides  g c,  so  that  </ o  =  295"  and  oc=  245",  we  di- 
vide c,  g^  in  same  proportion  at/,  and  carry  this  up  to  F  at  beam, 
which  is  the  point  of  greatest  deflection  of  beam,  and  is  distant  163" 
from  B,  and  197"  from  A.  We  scale//,  =  106"  (inch  scale)  and 
have  from  Formula  (97) 

rv  _  106.60.  288.  G500  _  ^  g^^,, 
^~~  27000000.1238  ""  ' 
1238  being  =  i,  the  moment  of  inertia  of  beam  as  found  in  Table 
XX.  The  beam  is  therefore  amply  stiff  even  to  carry  plastering. 
Irregular  Cross-  The  graphical  method  lends  itself  very  readily  to 
sections,  finding  centres  of  gravity  and  neutral  axes,  as  ex- 
plained in  the  chapter  on  arches,  and  also  for  finding  the  momenta 
of  inertia  of  difficult  cross-sections. 


MOMENT    OF    INERTIA. 


2G7 


If  we  have  an  irregular  figure  ABODE  (Figure  158)  we  divide 
it  into  simple  parts   I,  IT,  III  and  IV.     We  find   the  centres  of 

To  find  Neutral   gravity    g„    g,„ 
^*'^'  (Jul    and    <7,y    of 
each    part   and  draw   their   re- 
spective horizontal  neutral  axes 
through    these.       Anywhcre's 
make  a  line  a  e^  area  of  -whole 
figure  and  divide  it,  so  that: 
a&  =  area  of  I 
6c  =  area  of  II 
c(/  =  area  of  III  and 
de  =  area  of  IV. 
Select  pole  x  at  random,  draw 
xa,  xb,  xc,  xd,  and  x  e. 

From  any  point  of  horizontal 
-  p  g,  draw  fh  parallel  b  x  till  it  in- 

sects horizontal  <7„ ;  then  draw 
^^'  '^^*  /(_/ parallel  ex  to  horizontal//,,,; 

To  find  Moment  *^^°y^  parallel  rfr  to  last  horizontal,  and  finally  ko 
of  Inertia,    parallel  xe\  and  fo  parall"l  ax  till  they  intersect 

at  0.  A  horizon- 
tal through  0  is 
the  main  neu- 
tral axis  of  the 
vvhole.^  If  we 
multiply  the  area 
of  the  figure 
fohjh  by  the 
area  of  the  figure 
ABODE  {ho\h 
in  square  inches) 
■we  have  the 
value  of  moment 
of  inertia  i  oi 
ABODE  in 
inches,  around 
its  horizontal 
neutral  axis  o. 


«    r 


/cat* 


Fig.    159. 


»  The  point  of  intersection  of  this  line  with  a  main  neutral  axis,  found  similarlv, 
in  any  other  direction,  would  be  the  centre  of  gravity  of  the  whole  figure. 


268  SAFE   BUILDING. 

A  simple  way  of  obtaining  the  area  of  the  figure /o  k  would  be  to 
To  find  area.  draw  horizontal  lines  through  it  at  equal  distances 
beginning  with  half  distances  at  top  and  holtom,  and  to  multiply  the 
sum  of  these  horizontals  in  length  by  the  distance  apart  of  any 
two  horizontals,  all  measurements  in  inches.  This  will  approximate 
quite  closely  both  the  area  and  moment  of  inertia.  Of  course  the 
more  parts  we  take  in  all  of  the  processes,  the  closer  will  be  our 
result. 

A  practical  example  will  more  fully  illustrate  the  above. 

Example  VII. 

Rolled  Deck-  Find  horizontal  neutral  axis  and  the  corresponding 

beam,  moment  of  inertia  of  a  7"  —  55  pounds  per  yard  deck 
beam,  resting  on  its  flat  flange  {Figure  159). 

We  will  take  the  roll  as  one  part,  divide  the  web  into  four  equal 
parts,  the  flange  into  two  parts,  one  the  base  which  will  be  practi- 
cally rectangular,  and  its  upper  part  which  will  be  practically  tri- 
angular.    The  whole  area  we  know  is  for  wrought-iron  : 

55 
a  =  — =  5,  5  square  inches. 

The  bottom  rectangular  part  of  flange  will  be 
0'\-n^='^h  f  =  1>7  square  inches 
next  triangular  part 

AX    3 

The  web  parts 

5    ., 

^11  =  f'lii  =  «iv=  «v  =  — -: —  =  0,4  square  inches  each. 

Leaving  for  the  roll  at  top  a,  =  1,3 

We  now  make  the  horizontal  line  a  /t  =  5,5"  and  divide  it,  so  that 
a&=  1,3  inches 

bc  =  cd  =  de  =  ef=  0,4  inches 
fg  =  0,9  inches  and 
gh  =  1,7  inches. 
Select  X  at  random  and  draw  xa,  xb,  xc,  etc. 
Draw  the  horizontal  neutral  axes  T,  II,  III,  etc.,  through  their  re- 
spective parts.     Begin  anywhere  on  I  and  draw  j  k  parallel  &  a:  to 
line  II;  then  kl  parallel  ca:  to  III;  then  Im  parallel  dx  to  IV;  then 
JTjn  parallel  ex  toY;  then  np  parallel /x  to  VI ;  then p^  parallel 


HOI.LKD    DECK-BKAM.  2G9 

gx  to  VII;     Now  draw  from  7  the  line  qo  parallel  xh,  and  f rom  y 

„     .       ^  ,  the  line  jo  parallel  ax  till  thev  intersect  at  0.     A 

Horizontal  j       i  '        1        •       r       1     1 

Neutral  Axis,  horizontal  through  0  is    the  neutral  axis  of  whole 

beam.  We  will  now  make  a  new  drawing  of  figure  j  oq  for  the  sake 
of  clearness.  Draw  horizontals  through  it  every  inch  in  height  be- 
ginning at  both  top  and  bottom  with  one-half  inch.  The  top  one 
scales  nothing,  the  next  i",  then  f ,  then  U",  then  2|",  then  1^", 
Area  of  Dia-  ^"'^^  t^^^  bottom  one  ^",  the  sum  of  all  being  6^V' 
gram,  or  6,41G".  This  multiplied  by  the  height  of  the 
parts,  which  is  one  inch,  would  give  us,  of  course,  6,416  square 
inches  area.  Multiplying  this  area  by  the  area  of  the  cross-section 
of  deck  beam  5,5  square  inches,  we  should  have 

1  =  5,5.6,416  =  35,288. 
Moment  of  In-        I"  '^ablc  XX  it  is  given  as  35,  1  so  that  we  are 
ertiaof  Beam,  not  very  far  out. 

If  we  had  taken  more  parts,  of  course  the  result  would  have  been 
more  exact. 

Reducing  When   constructing   plate   girders  of   large  size, 

^"^^^'Cirdefs.  much  material  can  be  saved  by  making  the  flanges 
heaviest  at  the  point  of  greatest  bending-moment,  and  gradually  re- 
ducing the  flanges  towards  the  supports. 

This  is  accomplished  by  making  each  flange  at  the  point  of  great- 
est bending-moment  of  several  thicknesses  or  layers  of  iron,  the  outer 
layer  being  the  shortest,  the  next  a  little  longer,  etc.  Of  course  the 
angles,  which  form  part  of  the  flange  are  kept  of  uniform  size  the 
whole  length,  as  it  would  be  awkward  to  attempt  to  use  different 
sized  angles.  Generally  (though  not  necessarily)  the  inner  or  first 
layer  of  the  flange  plates,  is  also  run  the  entire  length.  Of  course; 
where  the  flanges  are  gradually  reduced  in  this  way,  it  becomes  ne- 
cessary to  figure  the  bending-moment  and  moment  of  resistance  at 
manv  points  along  the  plate  girder  to  find  where  the  plates  can  be  re- 
duced. This  would  be  a  wearisome  job.  By  using  the  graphical 
method,  however,  it  can  be  easily  accomplished.  Referring  back  to 
Fi<Ture  151,  we  take  the  point  of  greatest  bending-moment  (at  «?,)  of 
the  beam  A  B.  The  required  moment  of  resistance  at  this  point,  it 
will  be  remembered  was  the  length  (inch  scale)  of  vertical  S  through 
C  DE  FG.  We  now  decide  what  size  angles  we  propose  using  and 
settle  the  necessary  thickness  of  the  flanges  by  Formula  (36),  insert- 
ing for  the  value  of  r,  the  length  (inch  scale)  of  v  or  vertical  at  E. 
Further  a.  will,  of  course,  be  the  sum  of  the  area  of  two  angles,  d  the 


270  SAFK   BUILDING. 

total  depth  of  girder  in  inches  and  h  the  breadth  of  flange,  in  inches, 
less  rivet  holes.  The  above  is  on  the  assumption  that  the  distance 
xy  of  pole  X  from  load  line  d  a  was  equal  to  the  safe  modulus  of  rup- 
ture ( -J,]  of  steel  or  wrought-iron  according  to  whichever  material  we 
were  using,  or  we  should  have  : 

l"""'  (98) 


Thickness  of       ^ 

Flanges.  ""  b 

"Whei-e  2:=  the  thickness,  in  inches,  of  each  flange  of  a  plate 
girder  at  any  point  of  its  length. 

Where  v  =  the  length  of  vertical,  inch  scale,  through  upper  or 
resistance  figure,  providing  we  have  assumed  the  distance  of  pole 

from  load  line  (pound  scale)  =(~\oi  the  material. 

Where  d  =  the  total  depth,  in  inches,  of  the  plate  girder. 

Where  b  =  the  width,  less  rivet  holes,  in  inches,  of  the  flange. 

Where  a,  =  the  sum  of  the  areas  of  cross-section,  in  square 
inches,  of  two  of  the  angles  used. 

We  now  calculate  as  above,  the  thickness  x  of  flange  at  point  of 
greatest  bending- moment  and  then  decide  into  how  many  layers  or 
thicknesses  we  will  divide  the  flanges.     Say,  in  our  case  we  decide  to 

make  the  flange  of  four  layers  of  plates,  each  -  or  one  quarter  x  in 

thickness.     Then  make 

E,E,,  =  a,.d  (90) 

Where  E,  E„  =  the  amount  to  be  substracted  (inch  scale)  from 
moment  of  resistance  or  vertical  v  and  representing  the  work  of  two 
angles. 

Where  a,  =  the  sum  of  the  area  of  cross-section,  in  square 
inches,  of  the  two  angles. 

Where  d  =  the  total  depth,  in  iaches,  of  the  girder. 
Where  to  drop         Now  draw  through  E„  a  parallel  to  base  of  figure 

off  Plates,  c  G,  divide  ^„  E  into  as  many  parts  as  we  decide  to 
use  thicknesses  of  plates  (four  in  our  case)  and  draw  parallel  lines  to 
base  C  G  through  these  parts.  Vertically  over  the  points  Avhere 
these  lines  intersect  the  curve  or  outline  of  figure  C  D  E  F  G  will  be 
the  points  at  which  to  break  off  plates,  as  illustrated  in  drawing. 
This  method,  of  course,  is  approximate,  but  it  will  be  found  suffici- 
ently accurate  for  all  practical  purposes.     It  is  not  necessary  that  x 


KEDUCING    GIUDKK    FLANGES.  271 

or  E  7s„  be  divided  inlo  e(iii;il  parts.  Had  we  decided  to  use  plates 
of  varying  tliieknesst-s  we  should  simply  divide  E  Zi„  in  proportions 
to  corrospond  to  thicknesses  of  plates  in  their  proper  order,  heginnin<' 
at  i?„  with  plate  inunediatcly  next  to  angles  and  ending  at  7i  with 
extreme  central  outside  plate.  An  example,  more  fully  illustrating 
the  above,  will  be  given  in  the  chapter  on  jjlate  girders. 


,E    XII. 

.OOR-BEAMS. 

i  PER  Squakb  Foot  of  Floor 

] 

F  SPAN   IN     FKKT. 

Area  of  Section  per  square  foot 
of  floor  for  each  wood. 

O           i-H           CI           c«5           -.f           'O 

Hemlock. 

Spruce  or 
W.  Knc. 

White 
Oak. 

Georgia 
Pine. 

1 

1 

1 

1 

i 

1 

Inches. 
9 

9 

9 
10.3 
10.3 
10.3 
12 
12 
12 
12 

9 
12 
12 
13.7 
13.7 
10.3 
13.7 
16 
16 
12 
12 
16 
13.7 
16 
13.5 

Inclien. 

9 
10.3 

12 

12 

13.7 
16 

13.5 

Inclu'8. 

9 
10.3 

12 
12 

13.7 

16 

Inches. 

9 
10.3 

12 
12 

13.7 
16 

InchcB. 

1 

9 
10 

3 

1 

1 
—i — 

1 



1 

; 

1 

1 

— 1 — 
1 

1 

' 

1 

1 

1  ■■ 

1 

1 

1  - 

12 

12 

13.7 
16 

-i — 
1 

1 

1 

' 

1 

1 

T.IIiLE    XII. 

WOODEN    FLOOR-BEAMS. 


[Calc 

U..ATED 

FOB 

90  Pounds  pee 

Square  Foot  of 

Floor.] 

J 

M.lTETllAI,  Ol' 

i 

1 

s 

il 

si 

si 

I.ESGTII  OF  SPAN  IN    FKKT. 

< 

Area  of  Section  per  square  toot 
of  floor  for  each  wood. 

^    -    -    -.    -.    ^    -    ::    2    z    ^ 

! 

tmlocV  ^'•"""•l 

™° 

riMisiB 

,liichc«| 

3 
3 

3   1 

4 

4 
4 

IG 
10 
IG 
14 
14 
U 
12 
Hi 

o; 

i 

1 

1 

1 

1 

i 

i 

1 

i"' 

9 

9 
10.3 
10.3 
10.3 
12 
12 
12 
12 

9 
12 
12 
13.7 
13.7 
10.3 
13.7 
IG 
IG 
12 
12 
16 
13.7 
IG 
13.5 
13.5 
13.5 
15.4 
15.4 
15.4 
18 
18 
18 
18 
13..5 
18 
18 
20.0 
20.G 
15.4 
20.0 
24 
24 
18 
18 
24 
20.0 

18 

18 

24 

18 

20.6 

20.0 

20.0 

24 

24 

24 

24 

18 

24 

24 

27.4 

27.4 

20.G 

27.4 

32 

32 

24 

24 

22.5 

82 

22.3 

22,6 

25.7 

27.4 

25.7 

25.; 

SO 

34. 3 

82 

SO 

SO 

22.5 

9' 

10.3 

12 
12 

13.7 
10 

13.5 

l.i.4 

IS 
18 

20.0 
24 

18 

20.0 

24 
24 

27.4 

32 

22.5 

.■14.3 

9 
10.3 

12 
12 

13.7 
10 

13.5 
15.4 

IS 
18 

20.0 
24 

18 

20.0 

24 
24 

:    .4 
32 

22.5 

30 
30 

9 
10.3 

12 
12 

13.7 

IG 

13.5 
15.4 

18   ' 
18 

20.6 

24 

18 
20.6 

24 
24 

.. 
27.4 

32 
22.6 

2.1.7 

I,.cl»». 

9 

10.3 

12 
12 

13.7 
IG 

13.5 

16.4 

18 
18 

20.6 
24 

18 

2lMi 

24 
24 

27.4 

32 
22.5 

; 

j 

1 

[ 

6     5 

5    e 

0      7 
3     8 
S     9 
5     9 
5  10 
5  10 

5  11 
C 

0 
C 

6  2 
6     3 

6  3 
0     4 
0     5 
G     G 
0     G 
G      7 
Ij   10 

7  3 

7  10 
8 

8  1 
8     2 
8     i 
8     C 
8     7 
8     7 
8  10 
8  10 

8  11 
9 

9  2 
9     4 
9     4 
9     0 
9     8 
9     9 
9     9 
9  10 

10     3 
10     5 
10     8 

White  Oak, 
iieinlock, 
Spruce  or  \V.  P. 
White  Oal(, 
Hemlock, 
Hemlock, 
Spruce  or  W.  P. 
Spruce  or  W.  P. 
Georgia  Pine. 
White  Oak, 
White  Oak, 
Hemlock, 
Spruce  or  W.  P. 
Georgia  Pine, 
White  Oak, 
Elemlock, 
Spruce  or  W.  P. 
Georgia  Pine, 
Georgia  Pine, 
White  Oak, 
Georgia  Pine, 
Georgia  Pine, 
Hemlock, 
Spruce  or  W.  P. 
White  Oak, 
Hemlock, 
Spruce  or  W.  P. 
White  Oak, 
Hemlock, 
Hemlock, 
Spruce  or  W.  P. 
Spruce  or  W.  P. 
Georgia  Pine, 
White  Oak, 
White  Oak, 
Hemlock, 
Spruce  or  W.  P. 
Georgia  Piue, 
White  Oak. 
Hemlock, 
Spruce  or  W.  P. 
Georgia  Pine, 
Georgia  Pine, 
White  Oak, 
Georgia  Pine, 
Hemlock, 
Spruce  or  W.  P. 
Georgia  Pine, 
White  Oak, 
Hemlocli, 
Spruce  or  W.  P 
White  Oak, 
Hemlock, 
Hemlock, 
Spruce  or  W.  P 
Spruce  or  W.  P 
Georgia  Pine, 
White  Oak, 
White  Oak, 
Hemlock, 
Spruce  or  W.  P 
Georgia  Pine, 
White  Oak, 
Hemlock, 
Spruce  or  W.  P 
Georgia  Pine, 
Georgia  Pine, 
Hemlock, 
White  Oak, 
Spruce  or  W.P 
White  Oak, 
Hemlock, 
Georgia  Pine, 
Spruce  or  W.  P 
White  Oak, 
Hemlock, 
Hemlock, 
Georgia  Pine, 
Spruce  or  W.  P 
Spruce  or  W.  P 
Georgia  Piue. 

12 
12 
12 
12 
12 

in 

12 
Hi 
12 
12 
Hi 
IC 
10 
12 

ii: 
1(1 
h; 

12 
Hi 
Hi 
HI 
Hi 
IS 
IS 
18 
13 
18 
13 
18 
24 
18 
24 
13 
13 
24 
24 
24 
18 
24 
21 
24 
18 
24 
24 
24 
21 
24 
24 
24 
24 
24 
24 
24 
32 
24 
32 
24 
24 
32 
32 
82 
24 
32 
32 
32 
24 
32 
30 
32 
30 

iiO 
32 

3U 
40 
.'12 
30 
40 
30 

[ 

i 

1 

j 

3      ;         4 

3   ;     4 
3  1     i 

3  1     4 

4  1     4 
3         4 

, 

, 

. 

1 

, 

, 

' 

' 

1 

; 

, 

1 

1 

' 

1 

' 

; 

, 

1 

, 

12 

16 
16 
12 
IG 
14 
14 
14 
14 
12 
12 
12 
IG 
12 
14 
12 
16 
16 
16 
U 
14 
14 
12 
16 
12 
16 
16 
12 
IG 
14 
14 
14 
14 
12 
12 
!•! 

1 

, 

' 

1 

, 

1 

] 

1 

"^ 

, 

3 
3 
4 
4 
4 
3 
4 
4 

3 

4 
4 
3 
3 

3 
3 
3 
3 
4 
3 
4 
3 
3 
4 
4 
4 
3 
4 
4 
4 
3 
4 
4 

t 
3 
4 

a 

3 
3 
3 
3 
4 
3 
4 
3 
3 
4 
4 
4 
3 

4 
8 
4 
3 
4 
3 
3 
3 
4 

3 
3 

4 
3 
4 

1   3 

4 
4 

4 
4 
4 
4 
4 
4 
4 
4 
4 
4 
4 
4 
0 
G 
G 
G 

G 

C 
G 
C 
G 

a 

0 
G 
G 
0 
C 
G 
C 
C 
C 

8 
8 
6 
8 
8 
8 
8 
8 
8 
8 
8 

8 

8 
8 
8 
8 
S 
8 
8 
8 
8 
10 
8 
10 
HI 
10 

11) 
10 
10 
10 
8 
10 
10 
10 

■ 

1 

. 

1 

\  1 

I 

' 

' 

, 

1 

' 

; 

' 

■ 

j 

1 

, 

1 

1 

I 

' 

' 

1 

1 

1 

' 

] 

' 

1 

, 

^ 

, 

' 

, 

1 

' 

J 

' 

] 

, 

' 

' 

1 

\ 

j 

; 

' 

' 

[ 

'D^' 

' 

, 

[ 

\ 

, 

' 

1 

?  ' 

1 

, 

1 

I 

6  1 

1 

1 

I 

1 

, 

\' 

1 

1 

9t 

J 

1^ 

1 

1 

1 

1 

6 

1 

1 

] 

dj 

\, 

1 

1 

' 

, 

h 

' 

1 

df 

' 

1 

. 

[ 

?  ' 

1 

; 

* : 

' 

1 

1 

\' 

1 

1 

r 

' 

1 

1 

' 

r 

' 

' 

' 

^' 

' 

' 

' 

> 

1 

1 

\ 

1 

I 

I 

_\ 

1 

1 

16 
12 
14 
16 
16 
12 
16 
14 
14 
14 
12 
16 
12 
16 
16 
12 
16 
14 
14 
14 
14 
12 
12 
12 
16 
16 
12 
16 
16 
14 
14 
14 
14 
12 
14 
12 
12 
16 
16 

* 

! 

4 

,- 

, 

1 

\ 

, 

1 

\ 

_, 

, 

»? 

1 

1 

10   10 

\ 

' 

[ 

10   10 

{ 

' 

1 

10  11 

J 

\ 

|U      2 
U      4 

u    c 

'11     c 
U     9 
'11     9 

>?'._ 

[ 

^, 

, 

>> 

1 

4 

, 

; 

? 

1 

* 

1 

11   11 

^ 

1 

1 

1 

^  ' 

1 

'l2 

^J 

' 

i 

' 

1 

\ 

1 

12     b 

k 

, 

; 

* 

, 

) 

1^ 

13 

1                                                                   0  1 

1 

13 

1  ' 

j_ 

— 

9  ' 

' 

V 

r 

Q 

■ 

b 

1 

0 

. 

<! 

' 

k  ■ 

I'l      ' 

b' 

111 

V 

I 

g 

0 

^ 

14   1 



• 


Area  of  Section  per  square  foot 
of  floor  for  each  wood. 


Spruce  or     -Wliite    i  Georgia 
W.Pine.        Oak.     I     pine. 


Inches. 

30 

30 


34.3 


Inchea. 


25.7 


30 
30 


34.3 


• 


Table 

XIX   (Continued). 

1 

f 

MATRItlAI.  OF 

i. 

1 

1 

.11 
as 

LESGTH  OF  SPii'  IS  FEET 

-1 

Area  of  Section  per  square  foot! 
of  floor  for  each  wood.          1 

2     2     ^     5     2     S     S     SI     S5     ?.     S     g     S     S     S 

Ucmlock, 

wTnl' 

30 
30 

34.3 

40 
32 

36 
36 

41.1 
31.5 

48 
36 

42 
42 

36 

41.1 
56 

48 
48 

54.9 
64 

GcoiBl« 

15 
15 
16 

15     4 
15     7 
15     8 
15     8 

15  10 
16 

16  2 
16     8 
16     3 
16     3 
16     4 
16     5 

16  9 
17 

17  2 
17     3 
17     3 
17      7 
17      7 

17  11 
18 

18 
18 

18  1 
18     3 
18     5 
18     8 

18  8 
IS     9 
IS  11 
19 

19  1 
19     5 
19     6 
19      7 
19      7 
19     9 

19  10 

20  I 
20     1 
20     0 
20    e 
20     7 
20  U 

20  11 
21 

21 
21 

21  4 
21     6 
21     S 
21     9 
21   10 
21   10 

21  11 

22  2 
22     4 
22     8 
22     8 
22  10 

22  10 
23 

23 
23 

23  6 
23     6 

23  11 
24 

24 
24 
24 

24  7 

24  11 
25 

25  3 
25     4 
25  10 
26 

26 

20     S 
27     5 
29 

White  Oak, 

White  Oak, 

Hemlock, 

Spruce  or  W.  P. 

Georgia  Fine, 

White  Oak, 

Hemlock, 

Hemlock, 

Spruce  or  W.  P. 

Spruce  or  W.  P. 

Georgia  Pine, 

Georgia  Pine, 

White  Oak, 

Hemlock, 

White  Oak, 

Spruce  or  W.  P. 

White  Oak, 

Georgia  Pine, 

Hemlock, 

Hemlock, 

Spruce  or  W.  P. 

Spruce  or  W.  P. 

Georgia  Pine, 

White  Oak, 

White  Oak, 

Hemlock, 

Georgia  Pine, 

Hemlock, 

Spruce  or  W.  P. 
Georgia  Pine, 

Spruce  or  W.  P 

White  Oak, 

Hemlock, 
White  Oak, 

Elemlock, 
Spruce  or  W.  P. 
Spruce  or  W.  P. 
Georgia  Pine, 
Georgia  Pine, 
Wliite  Oak, 
White  Oak, 
Hemlock, 

llemlock. 
Spruce  or  W.  P. 
Spruce  or  W.  P. 
Georgia  Pine, 
Georgia  Pine, 

hemlock, 
White  Oak, 
White  Oak, 

lemlock, 
Spruce  or  W.  P. 
Spruce  or  W.  P. 
Georgia  Pine, 
White  Oak, 

Jemlocfc, 
Georgia  Pine, 
White  OakV 
Hemlock, 
Sprnce  or  W.  P. 
White  Oak, 
Spmee  or  W.  p. 
Georgia  Hne, 
Georgia  Pine, 
White  Oak. 

Jeralock, 
Hemlock, 
Spruce  or  W.  P. 
Spruce  or  W.  P. 
Georgia  Pine, 
White  Oak, 
White  Oak. 
Hemlock, 
Georgia  Pine, 
Spruce  or  W.  P. 
Georgia  Pine, 
White  Oak, 
Hemlock. 
Georgia  Pine, 
Spruce  or  W.  P. 
Georgia  Pine, 
Georgia  Pine, 
White  Oak, 
Georgia  Pine, 
Georgia  Pino, 

30 
40 
40 
40 
30 
40 
36 
40 
36 
40 
30 
40 
36 
36 
40 
36 
36 
40 
36 
48 
36 
48 
36 
36 
48 
48 
40 
42 
48 
36 
42 
48 
48 
42 
42 
48 
42 
36 
48 
48 
42 
42 
66 
42 
56 
48 
42 
48 
42 
56 
56 
48 
56 
48 
48 
48 
42 
56 
50 
48 
48 
30 
42 

-l.S 

4S 
64 
48 
43 
64 
64 
56 
04 
48 
64 
64 
56 
64 
48 
6t 
04 
C4 
04 

3 
4 
4 
4 
3 
4 
3 
4 
3 
4 
3 
4 
3 
3 
4 
3 
3 
4 
3 
4 
3 
4 
3 
3 
4 
4 
4 
3 
4 
3 
3 
4 
4 
S 

a 

4 
3 
3 
4 
4 
3 
3 
4 
3 
4 
4 
3 
3 
3 
4 
4 
3 
4 
4 
3 
3 
9 
4 
4 
3 
3 
4 
3 
4 
4 
3 
4 
3 
4 
3 
3 
4 
4 
4 
4 
3 
4 
4 
4 
4 
3 

4 
4 

10 
10 
10 
10 
10 
10 
12 
10 
12 
10 
10 
10 
12 
12 
10 
12 
12 
10 
12 
12 
12 
12 
12 
12 
12 
12 
10 
14 
12 
12 
14 
12 
12 
14 
14 
12 
14 
12 
12 
12 
14 
14 
14 
14 
14 
12 
14 
16 
14 
14 
14 
16 
14 
IS 
16 
16 
14 
14 
14 
16 
16 
jl 
It 
14 
11 
16 
16 
16 
16 
16 
10 
16 
16 
14 
16 
16 
16 
16 
14 
16 
16 
16 
16 
16 
16 

16 
14 
14 
14 
14 
16 
12 
16 
12 
12 
16 
16 
14 
12 
14 
14 
14 
12 
16 
12 
16 
16 
12 
16 
14 
12 
16 
14 
14 
16 
14 
12 
16 
14 
12 

12 

12 

12 

12 

12 

12 

1 

1 

1 

1 

1 

1 

I 

1 

1 

1 

1 

1 

1 

so"' 

30 

34.3 

34.3 

25.7 

34.3 

27 

40 

27 

40 

30 

SO 

27 

30.9 

40 

30.9 

S2 

34.3 

36 

36 

3IJ 

36 

27 

36 

30 

41.1 

40 

31.5 

41.1 

30,9 

31.5 

41.1 

48 

31.5 

S6 

86 

36 

36 

48 

36 

42 

42 

42 

42 

41.1 

31.5 

36 

42 

42 

48 

S6 

48 

48 

36 

41.1 

36 

4t( 

50 

41.1 

41.1 

56 

42 

42 

56 

48 

48 

48 

48 

36 

48 

48 

54.9 

48 

54.9 

41.1 

64.9 

64 

56 

64 

48 

48 

64 

54.8 

64 

34.3 

27 
40 

30.9 

36 
36 

41.1 
31.5 

48 
30 

42 
42 

36 
48 

41.1 

4» 
48 

54.9 
64 

Inches. 
31.3 

30.9 

36 
36 

41.1 
31.5 

•IS 
36 

42 
42 

36 

48 

41.1 
56 

48 
48 

54.9 
64 

25.7 

30 
30 

34.3 

27 

40 
30.9 

36 
36 

41.1 
31.5 

48 

36 

42 
42 

30 

48 
41.1 

56 

48 
48 

54.8 
04 

; 

1 

1 

1 

! 

' 

' 

( 

1 

1 

[ 

, 

1 

, 

' 

. 

, 

1 

' 

. 

' 

, 

1 

1 

1 

' 

■ 

1 

; 

1 

. 

— 

>  ' 

1 

1 

' 

1 

bl 

t 

1 

' 

' 

, 

1 

' 

T 

' 

, 

. 

, 

1 

1 

' 

, 

p. 

1 

1 

1 

1 

' 

r 

1 

i 

■ 

6. 

' 

1 

1 

Q 

1 

' 

, 

1 

Q 

1 

1 

, 

, 

1 

.1 

. 

' 

[ 

■ 

1 

' 

b| 

1 

?' 

' 

t' 

' 

i 

9 

1 

i^ 

' 

, 

^ 

?  ' 

, 

' 

<^   1 

, 

■  ' 

Y  ' 

' 

. 

' 

■ 

' 

' 

' 

. 

1 

' 

^ 

' 

' 

' 

; 

' 

' 

, 

• 

, 

9  , 

, 

' 

' 

1 

■ 

' 

1 

1 

1 

, 

' 

' 

■ 

. 

' 

1 

'    1    ' 

' 

1 

1 

, 

■ 

1 

' 

' 

1 

' 

' 

1 

■ 

' 

1 

; 

.   1 

' 

' 

9   1 

, 

, 

1 

1 

\ 

1 

1 

, 

■ 

* 

, 

V 

1 

1 

57 

> 

1 

b 

t  ' 

?  ' 

1 

K: 

^       "^ 

!> .  ) 

r 

; 

4: 

' 

X 

>? 

' 

1 

L 

*! ; 

1 

1 

V 

, 

1 

-c^, 

XIII. 

5RACED   SIDEWAYS. 


SPAN   IN   FEET. 


Ol  O  ^  <M  CO  -t  O  -O  t-.  X'  C5  O  .-I  (M  S^  -rt)  m 
I—  C^  C^l  O)  C^  fN  (M  C^  C-l  •>!  C^4  CO  IC  CO  CO  CO  CO 


Safe  Uniform  Load  in  Lbs. 
Foii  Full  Lines. 


Goorsia 
Pine. 


White 

Spruce  or 

Onk. 

W.  Pine. 

nemlock. 

500 

475 

445 

1000 

f)50 

890 

1500 

1425 

1335 

2000 

1900 

1780 

2500 

23  75 

2225 

3000 

2850 

2670 

3500 

3325 

3115: 

4000 

3800 

3560 

4500 

4275 

4005 

5000 

4  750 

4450 

5500 

5225 

4895 

6000 

5700 

5340 

/ 


Table   XIII. 
WOODEN    GIRDERS,  BRACED    SIDEWAYS. 


KlY, 

FOR  FLOORS. 


LENGTH  OF  SPAN  OF  BEAM  IN  FEET. 


»r  Iron  Beams  ;  but  length  of  span  in  feet  must  not  exceed  twice  the  depth 


^ 


Table  XIV. 
IRON   I  BEAMS   FOR   FLOORS. 


r&T 

,.»„!, .~  J 

ITS 

;;:,E 

''■'"■  1  •«•'"'   :  i  :  i  :  :""";"""1""~T:T'T:':":1": "":""" """""111''' 

"■  'i"ji 

1        113 

1    a 

1  9 
2 

2  3 
2     7 

2  10 

3  2 

3    n 

3  9 
4 

4  3 

4  10 

5  2 

5    r, 

5     9 
G 

G     3 
G      7 

e  10 

7     2 
7     .0 

7  9 
8 

8  3 
8      7 

1  '"s 

2 

2  4 

2  8 
3 

3  4 

3  8 
4 

4  4 
4     8 
5 

6     4 
.■j     8 
G 

6  4 
G     8 

7  4 

7  8 
8 

8  4 

8  8 
9 

9  4 
9     8 

10 

2  "' 

2     5 

2  10 

3  2 

3  7 
4 

4  5 

4  10 

r,    2 

5  7 

6  5 
G   10 

7  2 

7  7 
8 

8  S 

8  10 

9  2 
9     7 

10 

1,'°'c  *  ^^-^-  +  ^J^-^-fk--t-^f+■'-^•  +  •l-^■+  +  ^-^•■l-^-  +  +  +4•^t44■4^■4-r--t^■  +  +4  .4' 

I 

„           +    4    4  4  +,  1   .|     1-  !p  J-  1.^  1  ■(    fi     1  -1     1-  1-   1     (■  1-  H-  i-.t.   K   f    >    1     1    4    1  -1     1    -I-    1    4   -I-   i-  -1     1     t     I 

n 

1      5 

1  10 
2 

2  2 
2     3 
2     7 

2  10 
.1 

S     2 

3  !> 

3  7 
8  10 

4  2 

4  10 

a    2 

5  5 

r>    7 

a  10 

1  9 
2 

2  3 
2     S 

2  9 
3 

3  3 
3     G 

3  9 

4  3 

4    e 

4  9 
5 

0     3 

5  G 

5  9 
G 

6  3 
G     G 
G     9 
7 

7  3 
7     G 

^            4444;?4   4,V^t^S4gt^5  4:^44K4S^**£^•^4  4,stS^^4^.4  4   444, 

}  ; 

4            *    *    '^i-ffi25^;^Z4^ti<J4f  A4   4a(|tSS^4T_6i|?5:^4    4,44   4|^ 

1    G^+  *?^§04z§*i»s'^i?g*5*pfl2S5;;-r^>-?^+-.-4^4  e-J^r 

1     8 

1  10 
2 

2  2 
2     4 
2     (1 

2    a 

2  10 
3 

3  2 
3     4 
3     0 
3     8 

3  10 
4 

4  2 
4     4 
4     6 
4     8 
4  10 

5  ^--"|^t|'/4if4?'4gj?4r5P|^|:e.K'fir4  4  4j^.-t4,*-r4{r 

5     6^  +  :i('?W->^-|z:^ZtZ^^S*i^5?*?-i?P^^-  ,  j?^^^^^?.,4^5 

6      '  ^7^aiEdEir^Sf^2i^;iPli-£5£5''  -i*  ^  -  ^ ^^  ♦  '  i?;*^ 

6   6**^i.li:4'i  +  t,i*i^*'-?l^iZ'-*IS*^^»i:'^*;^**+4,^^** 

7      +  '-iM£Oit-§<-0'-^5Z;2^i47?^.  +  4^f..  .^.  ,,    ^..^^^ 

7  „.44ts-!r^tv^!?'Jj^t^g^t|r4T7Z^-47,»»4'r.^-t5',•^4Jij 

8      M*iM5*'^4<'-|Z*-ZZ^;K^*Z2^ii^*-'-<!!''*2i'^-i^*'-"F«'-^ 

?(i^|l,|i|Ot|5|^±,SW?4   4,U^4^^44    44<.-.^t...44.4 

9      -iIifEEf2  3E|j±|Z^2'^iL5^^'-^i:^*^-^+*4^;4  44444444^ 

9   6  ^Ht'^V'^*£'^U^^t'-HH*-^i'p^+-?^'-»^- -'-'-*-* -*-^«l 

,0      ♦EI  ti|*'Ii^-'2/'^^^i5i^*^Z-^'^^--*a*'- '■<-*■*-' *-+<••! 

„       r     t4t.||«.r|?4»*4»t2224444^44t^44#4»44f4»44T» 

?"*W||fl^tPi|f3f4|t..4/444;?'4,.Z,4....,^4^T4^ 

1-               ^.+     lL+^^luulf-ll   Jf*-    -**/    f   •*-*■*-.  -*     •■+     /  *    i-    -^    ^   i-     ^■^■*-^■^-^<■^-■^■^^-**     t- 

12        Cj^[.^J^    ■'■/r-|ff    r    *lH  '''    If   *"    i  /    ^     4---/++     t-4    4.    +    +-^^     i.    +    i.f^v-.     V    +    4.    +    4--V     t-i 

10  10 

11  2 
11      7 
12 

•^             f    -^    *-    1    4-1  t'ff+^    ^fiT'^f^r^^'^     <-    t     ''     |-y■*-■t-+/'-tt■-^t--•-^^■t-t'■'■^    +    ■»^■ 

u  "  * " '  1 H  Ifi  ♦  y  w  J  '■'  ^  y  / '  -  y + *  ►  '^  ^  ♦  -^  - '  —  -  ^  -  -  ^  *-  -  *-  --  -^ '- 

l-l      6,^^^^|^^j||'jJL  .^||i^  ^^+jJL^+i.^.^.^+^^++,^-,4^.  +  *-f.+.i.^  +  4.^^.+T. 

1^                                                                                                                                                                                      1 

For  Stool  Beama.  apace  oiie..i  i  irtar  .listanca  (li.'t^r.'^n  .-nnrr.'*)  liirgor  tbnn  for  Iron  Beams  ;  but  length  ot  Bpan  in  feet  milFt  not  exceed  twice  tbe  deptli 
of  beam  in  iucbes,  or  dellectiou  will  be  too  great  for  plaatyriiig. 

r 


XIX. 
TABLES   XX   TO    XXV. 


H    ARE    UXIFOUM    ABOVE    AND    KELOW   THE    XeUTUAL    AxIS. 


MANNKE   01<-    I.OADIXO. 


©a 


TO   OllTAIX  SAKK  LOAD 
IN   1'OrND.S. 


LENGTH  OF  SPAN 

NOT  TO  CUACK  PLASTERING 

MUST  NOT   KXCEED. 


---  L- 


■«',=r  W„ 

-"n-L 

or:  — 

ii\  -\-  ii\ 

V 

H-L 

ll\„z^lV„„  = 


2.  /. 


Wn,  +  "-n.,=-^ 


For  Iron.     I     For  Steel. 


GREATEST  ACTUAL 
DEFLECTION  WILL  BE. 


For  Iron. 


L==2.d      1  L  =  Vl.d     0=  .-^-. 

•*  b  i .  (I 


For  Steel. 


02^.  d  I 


8  = 


r|../ 


ibles  XX  to  XXV;  M=r  uniform  load  in  lbs.  ;  j«  =  centre  load  in  lbs. ;  «•,—.«•,,;  also  ji-,„  =  «•„,,  r=  couoen- 


■efuUv,  as  steel  varies  greatly  in  strength.     For  equal  deflections  of  steel  and  iron,  add 
le  transverse  values,  the  moduli  of  rupture  used  were:  for  iron,  120(»0  pounds  per  sc^uare 

as  follows  :  —  In  case  of  uniform  load  deduct  entire  weight  of   beam  ;    in  case  of   centre 
m  each  load  f  the  weight  of  beam;    in  case  of  loads  at  quarters  of    span,  deduct  from 

same  dimensions.     Tn  ordering  steel  or  iron,  give  either  the  required  dimensions  or  the 

ale  represent  the  following  Rolling  Mills :  — 

r>  — PHCENIX   IRON   COMPANY.   Philadelphia,   Pa. 

D  —  POTTSVILLEIRON   AND  STEEL  COMPANY,   Pottsville,   Pa. 

Pa.  F— PASSAIC  ROLLING   MILL  COMPANY,   Passaic,   N.  J. 

I 
;es  the  Mill  which  rolls  the  e.N^act  shape  given  in  the  Table;    the  other  letters  give  the 


TEE. 


Flanee 


Ranoe. 


DECK  BEAM. 


Taule   XIX. 
EXPLAIXINCt    USE    OF   TABLES    XX   TO   XXV. 


^ 


EEi!^ 


8=^. 


8=^ 


_^^ 


F^ 


^3 


!2J.rf 


'-  =  '^■"1^-0^1 


;  X  =  leiigt!iinf. 


J  given  la  Tables  XX  to  XXV ; 


Note.  —  If  the  transverse  values  (y),  given  for  steel,  are  used,  test  each  piece  carefully,  as  steel  varies  greatly  in  strength.  For  equal  deflections  of  steel  and  iron,  add 
only  7A%  to  iroa  transverse  values,  instead  of  35%  as  given  in  Tables.  In  calculating  the  transverse  values,  the  moduli  of  rupture  used  were:  for  iron,  12000  pounds  per  sfjuarQ 
inch,  and  for  steel,  15000  pounds  per  square  inch. 

From  the  "Safe  Load"  as  obtained  above,  tMucl  the  weight  of  beam  in  pounds,  as  follows:  —  In  case  of  uniform  load  deduct  enfiVc  weight  of  beam;  in  case  of  centre 
load,  deduct  one-half  the  weight  of  beam;  in  case  of  loads  at  thirds  of  span,  deduct  from  each  load  |  the  weight  of  beam;  in  ease  of  loads  at  quarters  of  span,  deduct  from 
each  load  J  tlio  weight  of  beam. 


■  (about  1%)  than  iro 


cxactlv  the 


In  onlei 


steel  . 


Steel  sections  will  be  slightly  ii 
pijuired  weiglit  of  section,  hcct  hoih. 

Tlie  capital  letters  in  the  lirst  column  Iicadod  "Mills  Rolling  Shape"  in  each  Table  represent,  tlip  following  Rolling  Mill: 

on.    N.  J 


n,  give  either  the  requiretl  dir 


■  the 


R  — PHCENIX   IRON  COMPANY,   Philadelphia,  Pa. 
I)_POTTSVILLE  RON  AND  STEEL  COMPANY.  Pottsville.  Pa. 
F— PASSAIC  ROLLING   MILL  COMPANY,  Passaic,   N.  J. 


A  — NEW  JERSEY  STEEL  AND  IRON  COMPANY,  T 

C— PENCOYD  IRON  WORKS,   Philadelphia.   Pa. 

K— UNION   IRON   MILLS  and   HOMESTEAD  STEEL  WORKS,   Pittsburgh,  Pa. 

In  the  columns  hcadtsd  "Mills  Rolling  Shape"  the  drst  letter  on  each  line  indicates  the  Mill  which  rolls  tlie  e.\;iLt  shape  given  in  the  Table;    the  other  letters  give  the 
Mills  which  roll  an  approximately  similar  shape. 


In  gel 


of  parts  of  i 


,  thev 


!  taken  as  shown  below 


r 


XX. 

D   STEEL   I   BEAMS. 

OF  THIS  Table,  see  Table  XIX.) 


al  to  Web. 


n J N 


:.6 

.5 
.5 

! 
► 
!.5 

5 
^4 

5 
'.3 
t.5 
I 

7 
1.7 

r.i 

1.9 
.3 

^.2 
5 
3 
4 
2 

r.5 

[.7 
),2 
1.7 

.3 
,5 

,4 
t.9 
9 
1.6 

L 


5«0 


60.67 
60.42 
61.99 
59.68 
34.80 
35.32 
35.15 
32.52 
31.24 
34.70 
31.47 
35.20 
35.76 
34.40 
23.32 
22.60 
23.35 
18.92 
22.46 
22.60 
23.98 
21,80 
24.22 
17.49 
17.77 
17.39 
18.42 
13.91 
15.52 
15.35 
16.40 
12.63 
11.70 
12.23 
11.83 
13.10 
13.16 
13.41 
8.64 
10.44 


Transverse  Value 
(f)  in  lbs. 


For  Iron.      For  Steel. 


1320000 
1160000 
990000 
917000 
552000 
748000 
460000 
864000 
800000 
740000 
655000 
563000 
556000 
458000 
508000 
504000 
376000 
454000 
510000 
377000 
375000 
290000 
306000 
356000 
282000 
251000 
250000 
300000 
278000 
258000 
237000 
336000 
282000 
268000 
208000 
211000 
199000 
167000 
181000 
168000 


1650000 
1450000 
1238000 
1146000 
690000 
935000 
575000 
1080000 
1000000 
925000 
819000 
704000 
695000 
573000 
635000 
630000 
470000 
567000 
637000 
471000 
409000 
363000 
382000 
445000 
353000 
314000 
312000 
375000 
347000 
322000 
297000 
420000 
353000 
335000 
260000 
264000 
249000 
209000 
226000 
210000 


Axis  Parallel  to  Web. 


«—■ H=^-H 


46.50 

51.78 
26.62 
31.50 
15.29 
27.46 
11.64 
40.84 
29.90 
33.79 
20 

18.34 

16.91 

13.13 

25.41 

20.90 

11.54 

15.50 

24.08 

12.98 

16.76 

8.74 

11.66 

15.80 

9.43 

8.01 

8.09 

11.30 

10.64 

11.08 

8.09 

23.16 

14 

11.23 
7.14 
8.44 
7.35 
4.92 
6.96 
7.55 


13.78 
16.57 

8.87 
10.08 
6.12 
9.55 
4.66 
13.90 
10.29 
12.12 
7.50 
7.34 
6.00 
5.34 
9.24 
7.96 
4.82 
6.09 
8.76 
5.46 
6.10 
4 

4.44 
6.32 
4.19 
3.53 
3.59 
4.74 
4.60 
4.79 
3.69 
8.62 
5.67 
4.99 
3.30 
3.80 
3.27 
2.46 
3.25 
3.35 


1.71 

2.16 

1.33 

1.04 

1.02 

1.37 

.94 

1.62 

1.25 

1.69 

1.02 

1.22 

1.17 

1.05 

1.52 

1.23 

.93 

.86 

1.42 

1.04 

1.43 

.87 

1.23 

1.18 

.90 

.85 

.91 

.83 

.96 

1.05 

.90 

1.54 

1.02 

.91 

.72 

.92 

.86 

.70 

.67 

.93 


Transverse  Value 
(r)  in  lbs. 


110000 
132600 
71000 
80600 
49000 
76000 
37000 
111200 
82300 
97000 
60000 
58700 
52800 
42700 
74000 
63700 
38500 
48700 
70100 
43700 
49000 
32000  I 
35500 
50500 
33500 
28200 
28700 
37900 
36800 
38300 
29500 
69000 
45400 
40000 
26400 
30900 
26000 
19700 
26000 
96800 


137800 
165700 
88700 
100800 
61200 
95500 
40600 
139000 
103000 
121000 
75000 
73400 
66000 
53400 
92400 
80000 
48200 
61000 
87600 
54600 
61000 
40000 
44400 
63200 
41900 
35300 
35900 
47400 
46000 
48000 
36900 
86000 
56700 
49900 
33000 
38600 
32700 
24600 
32500 
SS.'inO 


Table  XX. 

LIST   OF   IRON  AND    STEEL   I    BEAMS. 

(For  IsffoaMATiox  as  to  thk  Use  of  this  Table,  sbc  Table  XIX.) 


L 
P 

1 

J 

.Is- 

f 

is 

•3 

I 

i 

i 

r 

A.,.^ 

—     "-I--   1 

A.,. 

P„.„e, 

..web 

"--■(MF-  ! 

jr 

Ir 

P^ 

'("  tli'lb..'""  1 

tL 

11^ 
P 

P 

0)  in 

1bl'.""° 

Fo,In.«. 

F„»,„l. 

r„,w 

ro,s,«. 

20 
20 
20 
20 

lit 

\'} 
15 
15 
16 
15 
15 
16 

12 
12 
12 
12 
12 
12 
12 
12 
lOJ 
lOi 
lOJ 
'"i 

10 
10 
10 
10 

9 

9 

9 

9 

9 

9 

9 

8 

e 

8 

7 
7 
7 
7 
7 
6 
G 
6 
C 

e 

5 
5 

3 
3 
3 
3 
U 

272 
240 
200 
192 
160 
200 
125 
250 
240 
200 
195 
1,50 
145 
125 
170 
170 
125 
180 
170 
125 
120 
100 

96 
135 
105 

95 

90 
135 
112 
105 

90 
150 
135 
125 

99 

90 

85 

70 
105 

SO 

05 

75 

69 

C5 

CO 

55 

62 
120 

90 

64 

60 

40 

40 

36 

34 

30 

37 

30 

28 

24 

18 

27 

23 

21 

17 
M 

6.75 

7 

6 

6.25 

5 

5.75 

5 

5.875 

5.81 

5.56 

5.33 

5 

5.125 

5.50 

5.26 

4.79 

5.09 

5.50 

4.76 

5.50 

4.44 

5.25 

5 

4.50 

4.54 

4.60 

4.77 

4.625 

4. 025 

4.376 

5.375 

4.94 

4.50 

4.33 

4.375 

4.50 

4 

4.29 

4.60 

4 

3.91 

4 

3.81 

3.50 

3.75 

3.61 

5.25 

5 

3. 46 

3.60 

3 

3 

3 

2.844 

2.75 

3 

2.76 

2.76 

2.25 

2 

2.52 

2.60 

2.32 

2.25 

1.50 

0.69 
0.60 
0.50 
0.50 
0.50 
0.60 
0.42 
0.876 
0.93 
0.625 
0.77 
0.47 
0.44 
0.44 
0.6U 
0.66 
0.47 
0.96 
0.69 
0.49 
0.39 
0.44 
0.31 
0.47 
0.S75 
0.41 
0.31 
0.77 
0.50 
0.50 
0.S4 
0.60 
0.75 
0.67 
0.53 
0.41 
0.375 
0.30 
0.79 
0.375 
0.30 
0.53 
0.375 
0.44 
0.40 
0.30 
0.234 
0.625 
0.50 
0.46 
0.30 
0.25 
0.31 
0.30 
0.31 
0.25 
0.81 
0.25 
0.25 
0.81 
0.19 
0.39 
0.26 
0.19 
0.166 
0.126 

7.87 
G.63 
5.65 
5.09 
4.34 
6.44 
3.37 
7.22 
6.29 
6.06 
4.84 
4.47 
4.48 
3.35 
5.47 
5.50 
3.78 
4.26 
6.77 
3.80 
3.87 
2.71 
3.09 
4.84 
3.07 
2.96 
3.11 
3.67 
3.01 
3.24 
3.14 
5.59 
4.22 
4.41 
2.85 
3.07 
2.93 
2.40 
2.74 
2.82 
2.21 
2.29 
2.47 
2.08 
1.88 
1.90 
1.92 
4.78 
3.31 
1.61 
1.77 
1.42 
1.86 
1.20 
1.06 
.99 
1.40 
1.08 
1.07 
.71 
.68 
.93 
.86 
.85 
.68 
.19 

11.46 
10.64 
8.67 
9.02 
6.36 
7.14 
5.62 
10.56 
11.42 
7.88 
9.82 
6.06 
6.59 
6.80 
6.83 
6.88 
4.77 
9.48 
6.46 
4.90 
3.99 
4.53 
3.28 
3.68 
3.10 
3.58 
2.68 
6.16 
3.95 
4.02 
2.70 
3.82 
6.00 
3.61 
4.20 
2.93 
2.64 
2.20 
6.02 
2.30 
1.96 
2. 92 
1.90 
2.42 
2.24 
1.70 
1.30 
2.28 
2.08 
2.18 
1.37 
1.17 
1.18 
1.20 
1.26 
1. 01 
.86 
.76 
.76 
.99 
.61 
,84 
.53 
.40 
.35 
-14 

27.20 
24 

19.97 
19.20 
16.04 
20.02 
12.36 
25 
24 
20 

19.60 
15 

14.56 
12.50 
16.77 
17 

12.33 
18 
17 

12.60 
11.73 
10 

9.46 
13.36 
10.14 
9.50 
8.90 
13.50 
11.17 
10.50 
9.04 
15 

13.60 
12.33 
9.90 
9.07 
8,50 
7 
10.50 
8.03 
6.37 
7.50 
6.90 
6.58 
6 

5.50 
6.14 
11.84 
8.70 
5,40 
4.91 
4.0! 
3.90 
3.60 
3.38 
2.99 
3.66 
2.91 
2.90 
2.40 
1.77 
2.70 
2.26 
2.10 
1.71 
.52 

1650.3 
1460 
1238 
1146 
523.6 
707.1 
434.6 
813 
750 
694 
614 
523 
621.2 
430 
891.2 
385 
288 
340 
381.9 
282.6 
281.3 
218 
229.2 
233.7 
185.6 
165 
164 
187 
173.6 
161 
148.3 
189.1 
159 
150.8 
117 
118.8 
111.9 
93.9 
90.4 
83.9 
67.4 
54.3 
55.7 
49.8 
46 
44.8 
43.1 
64.9 
49.8 
28.4 
29 

23.5 
15.4 
14.9 
13.4 
12.1 
9.2 
7.6 
7.7 
5.6 
4.6 
3,64 
3.29 
3.09 
2.66 
.135 

105 
146 
123.8 
114.6 
69 

93.5 
67.6 
108 
100 
92.6 
81.5 
70,4 
69.6 
67.3 
63.6 
63 
47 
56.7 
63.7 
47.1 
46.9 
36.3 
38.2 
44.5 
35.3 
31.4 
31.2 
37.5 
34.7 
32.2 
29.7 
42 
35.3 
33.5 
26 

26.4 
24.9 
20.9 
22.6 
21 

16.9 
15.6 
15.9 
14.2 
12.9 
12.7 
12.3 
21.6 
16.6 
9.6 
9.6 
7.8 
6.1 
5.96 
5.4 
4.8 
4.6 
3.75 
3.84 
2.80 
2.25 
2.36 
2.16 
2.06 
1.77 
.21 

60.67 

60.42 

61.99 

59.68 

34.80 

36.32 

35.15 

32.52 

31.24 

34.70 

31.47 

35.20 

35.76 

34.40 

23.32 

22.60 

23.35 

18.92 

22.46 

22.60 

23.98 

21,80 

24,22 

17,49 

17,77 

17.39 

18.42 

13.91 

16.52 

15.35 

16.40 

12.03 

11.70 

12.23 

11.83 

13.10 

13.16 

13.41 

8.64 

10.44 

10.58 

7.24 

8.08 

7.56 

7.60 

8.0.5 

8.35 

5.48 

5.72 

5.29 

5.90 

5.86 

3.95 

4.14 

3.96 

4.04 

2.61 

2.57 

2.65 

2.33 

2.54 

1.32 

1.46 

1.46 

1.56 

.26 

1320000 

1160000 

990000 

917000 

562000 

748000 

460000 

864000 

800000 

740000 

656000 

663000 

556000 

458000 

608000 

604000 

376000 

464000 

610000 

377000 

375000 

290000 

306000 

356000 

282000 

251000 

260000 

300000 

278000 

258000 

237000 

336000 

282000 

268000 

208000 

211000 

199000 

167000 

181000 

168000 

135000 

124000 

127000 

U3800 

103200 

101000 

98500 

172000 

132000 

76000 

76800 

62400 

4S800 

47700 

42800 

38400 

36800 

30000 

30700 

22400 

18000 

18900 

17300 

16500 

14200 

1650000 

1450000 

1238000 

1146000 

690000 

935000 

676000 

1080000 

1000000 

925000 

819000 

704000 

696000 

573000 

635000 

630000 

470000 

567000 

637000 

471000 

469000 

363000 

382000 

446000 

363000 

314000 

312000 

375000 

347000 

322000 

297000 

420000 

353000 

836000 

260000 

264000 

249000 

209000 

226000 

210000 

169000 

155000 

169000 

142000 

129000 

127000 

123000 

216000 

160000 

95000 

96000 

78000 

61000 

59600 

63500 

4S000 

46000 

37600 

,38400 

28000 

22500 

23000 

21600 

20600 

17700 

2160 

46.50 
51.78 
26.62 
31.60 
16.29 
27.46 
11.64 
40.84 
29.90 
33.79 
20 

18.34 
16.91 
13.13 
25.41 
20.90 
11.54 
15.50 
24.08 
12.98 
16.76 
8.74 
11.66 
15.80 
9.43 
8.01 
8.09 
11.30 
10.64 
11.08 
8.09 
23.16 
14 

11.23 

7.14 

8.44 

7.35 

4.92 

6.96 

7.55 

4.56 

4.87 

5.42 

4.16 

3.15 

3.90 

3.43 

18.59 

10.78 

2.61 

2.74 

1.61 

1.G8 

1.74 

1.21 

1.04 

1.74 

1.11 

1.17 

.58 

.31 

.84 

.77 

.66 

.48 

.069 

13.78 
16.57 
8.87 
10.08 
6.12 
9.66 
4.66 
13.90 
10.29 
12.12 
7.50 
7.34 
6.60 
6.34 
9.24 
7.96 
4.82 
6.09 
8.76 
5.46 
6.10 
4 

4.44 

6.32 

4.19 

3.53 

3.69 

4.74 

4.00 

4.79 

3.69 

8.62 

5.67 

4.99 

3.30 

3.86 

3.27 

2.46 

3.25 

3.85 

2.27 

2,50 

2.71 

2.18 

1.80 

2.08 

1.90 

7.08 

4.32 

1.45 

1.57 

1.07 

1.12 

1.16 

.85 

.76 

1.16 

.81 

.85 

.62 

.31 

.67 

.62 

.47 

.43 

.092 

1.71 

2.16 

1.33 

1.64 

1.02 

1.37 
.94 

1.62 

1.25 

1.69 

1.02 

1.22 

1.17 

1.05 

1.52 

1.23 
.93 
.86 

1.42 

1.04 

1.43 
.87 

1.23 

1.18 
.90 
.85 
.91 
.83 
.96 

1.05 
.90 

1.54 

1.02 
.91 
.72 
.92 
.86 
.70 
.67 
.93 
.71 
.66 
.786 
.62 
.53 
.71 
.67 
1.57 
1.24 
.46 
.66 
.40 
.43 
.483 
.36 
.35 
.48 
.38 
.40 
.22 
.175 
.31 
.35 
.30 
.28 
.013 

110000 

132600 

71000 

80600 

49000 

76000 

37000 

111200 

82300 

97000 

60000 

68700 

62800 

42700 

74000 

63700 

38600 

48700 

70100 

43700 

49000 

32000 

35500 

50500 

33500 

28200 

28700 

37900 

36800 

38300 

29500 

69000 

46400 

40000 

26400 

30900 

26000 

19700 

26000 

20800 

18200 

20000 

21680 

17400 

14400 

10600 

15200 

66600 

34500 

11600 

12600 

8500 

9000 

9260 

6800 

6000 

9500 

6500 

6800 

4160 

2500 

6360 

4960 

3760 

3440 

137800 

165700 

88700 

100800 

61200 

95500 

46600 

139000 

103000 

121000 

75000 

73400 

66000 

53400 

92400 

80000 

48200 

61000 

87600 

54000 

61000 

40000 

44400 

63200 

41900 

35300 

35900 

47400 

46000 

48000 

36900 

86000 

66700 

49900 

33000 

38600 

32700 

24600 

32500 

33500 

22700 

25000 

27100 

21800 

18000 

20800 

19000 

70800 

43200 

14500 

15700 

10700 

11200 

11580 

8600 

7600 

11600 

8100 

8600 

6200 

3100 

6700 

6200 

4700 

4300 

920 

A,B 

E 

D,B.C,E.... 

D.B.E 

D.B.E 

A 

B.C.D 

R,I),E 

A,C 

A,U,E 

A.B.C.D.E.F. 
A,B,C,D,F... 

A.B.C.D.F... 

C,D,E 

A,B,D,F 

A,1J,C,D.E,F, 

A.B.CD.F... 
A,B,C,D,E,F. 

A,B,D,E  .... 

A.B.CD.F... 
A  B,C,D,E,F. 
A,D,E,F 

A  B.C.D.E.F, 

A,F 

A,B,D,E,F  .. 

A.B.C.D.F  .. 

r 


Tablk    XXI. 

r   AND    STEEL   CHANNELS. 

0  Use  of  this  Tabi.k,  skk  T.abi-k  XIX.) 


Ixis  Normal        "T^ 
to  Web.    "     J 

1   Axis  Parallel        .    ^ 

!     to  Web.     " g***^ " 

>  . 

*  V 

Transverse 

0  0 

'~^_. 

Z  '-^ 

Trjiiisverso 

i'ls 

Hi 

s  c==--r  Value  (v)  in  lbs 

lie 

Sjs'ta 

vis 

Valine  (/•)  in  lbs. 

»-< 

3  5 

CO 

For  Iron 

.  For  Steel 

1- 

£  '5  ^ 

:»«5' 

«ll= 

For 
Iron. 

For 
Steel. 

J4.0 

70.0( 

)  29.20 

560000 

) 700000 

21.60 

6.93 

1.21 

1.06 

55600  69500 

)7..5 

62.3f 

)  31.30 

49840( 

) 623000 

19.70 

6.84 

1.32 

1.09 

54720  68400 

;o.7 

88.0f 

>  28.73 

70500C 

881000 

3  7.56 

10.05 

1.63 

1.28 

8050< 

100).JOO 

i4.G 

7;{.9^ 

27.73 

59150C 

739400 

23.61 

7.15 

1.18 

1.08 

5720f 

71500 

SCO 

78.13 

30.84 

62500C 

781300 

32.25 

9.24 

1.70 

1.26 

74O0f 

92JOO 

(7.0 

70.2(j 

30.03 

562080 

702600 

31.41 

8.92 

1.80 

1.23 

7136( 

89200 

9.1 

59.88 

29.94 

479000 

598800 

18.27 

6.09 

1.22 

1.00 

48720 

60900 

6.0 

50.13 

31.33 

401000 

501300 

14.47 

4.74 

1.21 

0.95 

38000 

4  7400 

9A 

55.41 

19.07 

443300 

554100 

20.65 

6.55 

1.16 

1.16 

5250)0 

65500 

1.6 

47.64 

20.83 

381000 

476400 

17.87 

6.20 

1.28 

1.12 

496O0 

62000 

4.3 

35.00 

21.40 

280000 

350000 

8.93 

3.68 

0.88 

0.86 

29440 

36800 

8.9 

27.70 

21.10 

221600 

277000 

5.44 

2.39 

0.67 

0.73 

19120 

2390)0 

3.2 

25.03 

21.89 

200000 

250300 

5.04 

2.24 

0.72 

0.755 

18000 

22400 

5.7 

39.29 

15.72 

314200 

392900 

8.44 

3.13 

0.56 

0.80 

250)40 

31300 

1.5 

30.25 

20.16 

242000 

302500 

7.11 

3.29 

0.79 

0.84 

26320 

32900 

3.7 

20.62 

20.79 

164960 

206200 

3.22 

1.62 

0.55 

0.62 

12960 

1620)0 

9.4 

24.64 

12.32 

197000 

246400 

4.96 

1.98 

0.47 

0.69 

16000 

1980)0 

6.0 

22.10 

15.50 

176800 

221000 

5.20 

2.55 

0.71 

0.80 

20400 

25500 

8.4 

16.84 

14.73 

134700 

168400 

3.84 

1.81 

0.64 

0  63 

14500 

18100 

7.0 

16.60 

15.90 

132800 

166000 

3.20 

1.70 

0.58 

0.70 

13600 

1  7000 

0.0 

28.00 

10.82 

224000 

280000 

7.79 

2.93 

0.61 

0.84 

23440 

29300 

8.6 

25.72 

11.59 

205760 

257200 

5.26 

2.35 

0.47 

0.76 

18800 

23500 

0.5 

20.10 

10.92 

161000 

201000 

3.02 

1.31 

0.33 

0.626 

10500 

1310)0 

7.4 

19.4  7 

12.98 

155760 

194700 

3.51 

1.78 

0.47 

0  66 

14240 

1 7800 

3.2 

14.63 

12.84 

117040 

146300 

1.97 

1.15 

0.35 

0.53 

9200 

11500 

4.0 

12.80 

13.33 

102400 

128000 

2.20 

1.14 

0.46 

0.565 

9100 

11400 

8.7 

24.16 

9.97 

193300 

241600 

7.30 

2.75 

0.67 

0.91 

22000 

2750)0 

9.1 

19.80 

9.92 

158400 

198000 

4.25 

2.02 

0.47 

0.73 

16160 

20200 

2.1 

18.24 

11.73 

146000 

182400 

5.35 

2.35 

0.76 

0.85 

18800 

23500 

8.8 

13.07 

11.76 

104600 

130700 

2.53 

1.35 

0.51 

0.63 

10800 

13500 

2.0 

9.33 

11.42 

74640 

93300 

1.52 

0.99 

0.41 

0.55 

7920 

9900 

4.5 

16.14 

7.67 

129100 

161400 

4.00 

1.98 

0.48 

0.73 

15840 

19800 

0.1 

15.03 

8.12 

120200 

150300 

3.53 

1.69 

0.48 

0.755 

13500 

16900 

8.4 

12.10 

7.81 

97000 

121000 

1.94 

0.98 

0.31 

0.584 

7840 

9800 

4.5 

11.12 

9.89 

89000 

111200 

2.54 

1.46 

0.56 

0.76 

11700 

14600 

8.2 

7.06 

9.55 

56480 

70600 

1.06 

0.71 

0.36 

0.50 

5680 

7100 

7.9 

10.83 

6.30 

86640 

108300 

2.85 

1.53 

0.48 

0.68 

12240 

15300 

5.6 

7.31 

5.5  7 

58500 

73100 

1.11 

0.62 

0.24 

0.51 

4960 

6200 

7.1 

7.74 

7.53 

62000 

77400 

1.96 

1.10 

0..54 

0.715 

8800 

11000 

7.6 

5.03 

7.05 

40240 

50300 

0.75 

0.49 

0.30 

0.470 

3920 

4900 

8.0 

9.33 

4.24 

74600 

93300 

2.95 

1.43 

0.45 

3.788 

11440 

14300 

3.4 

7.80 

4.33 

62400 

78000 

1.84 

0.95 

0.34 

3.662 

7600 

9500 

1.7 

7.23 

4.82 

58000 

72300 

2.12 

1.20 

0.47 

3.725 

9600 

12000 

7.2 

5.73 

5.21 

46000 

57300 

1.30 

0.80 

0.39 

3.630 

6400 

8000 

0.4 

3.47 

4.74 

27760 

34700 

0.62 

0.46 

0.28  ( 

3.400 

3680 

4600 

3.4 

5.34 

3.17 

42720 

53400 

1.50 

0.94 

0.36 

X640 

7520 

9400 

0.3 

4.12 

3.03 

33000 

41200 

m 

0.44 

0.19  ( 

3.493 

3520 

4400 

A,F. 
F,A. 
F,A. 


A,F. 

F,A. 
A,P. 


A,B,C,D,E . 
B,A,C,D,E . 
B,A,C,U,E . 

A,D,E 

A 

E,A,B,C.... 
A,B,C,D,E,F.  1 
A,B,C,D,F.. 


A,C,E 

A,C,D,E . . . 
A,B,C,D,E  . 

C,B,D 

E,C 

A,B,C,D.E  . 
A,B,C,D,E  . 
B,A,C,1)  .  . . 

A 

A,B,C,F.... 
A,C,D,E,F  . 
A,B,C,D,E,F. 
B,A,C,D,E,F.  I 

E,C 

A,C,D,E  . . . 
D,A,C,E  . . . 

B,D,F 

U.C.E 

B.A'.Cn.E. 

A.B.I) 

I' 


o.so 

".!■<•> 

1. 00 

HM 

ItMl 

a.fti 

11.46 

S.liO 

0..19 

■?.m 

0.3.1 

1.(19 

1 .00 

a.M 

0.44 

■?..n 

0.2« 

1.44 

O.HO 

l..iO 

0.4  a 

l.SO 

0..')75 

l.liO 

0.30 

1.30 

l.Oli 

1.7S 

0.875 

1.83 

1.17 

HM 

s.oo 

u;h,;i 

.1.62 

7.00 

l.iS.2 

(i.ai. 

lii.OO 

233.7 

4.l!( 

9.00 

181.5 

3.0B 

5.94 

12S.7 

7..'iO 

10.50 

129.4 

.3.90 

7.50 

U6.0 

3.60 

c.ou 

HK.4 

:■»" 


0.545 

1.03 

0.60 

0.65 

0.25 

0.80 

0.19 

0.44 

0.44 

0.91 

0.31 

0.71 

0.20 

0.60 

0.17 

0.4? 

0.53 

0.67 

0.376 

0.48 

0.20 

O.flS 

0.26 

(1.34 

0.27 

0.27 

0.13 

O.U 

117040 
102400 
193300 
158400 
146000 
104600 
74640 
129100 
120200 
97000 
89000 
56480 
86640 
.58500 
62000 
40240 
74600 
62400 
68000 
46000 
27760 
42720 
33000 
30400 
20320 
28000 
22160 
15600 
12960 
14400 
12080 
IU640 
5680 
3840 
1520 


-liUlO 

14.47 

4.74 

1.21 

0.95 

380(111 

.17100 

.  tlud 

20.65 

6.55 

1.16 

1.16 

526(1(1 

ll.i.'iOO 

17.87 

6.20 

1.28 

1.12 

49600 

*;■_'(  1(10 

(t 

8.93 

3.68 

0.88 

>.»6 

294411 

IldSllO 

) 

5.44 

2.39 

0.67 

1.73 

19120 

23900 

5.04 

2.24 

0.72 

).7.66 

18000 

22400 

8.44 

3.13 

0.66 

1.80 

25040 

31300 

_■   ,H(I 

7.11 

3.29 

0.79 

1.84 

26320 

32900 

.Jnll 

3.22 

1.62 

0.55 

).62 

12960 

16200 

',  luo 

4.96 

1.98 

0.47 

1.69 

16000 

19800 

jinul) 

5.20 

2.65 

0.71 

1.80 

204011 

25500 

'    .  KM) 

3.84 

1.81 

0.64 

163 

14600 

18100 

.i.UUd 

3.20 

I.!0 

0.68 

1.70 

I36O0 

17000 

-, 1,1(1(1 

7.79 

2.93 

0.01 

1.84 

23440 

29300 

7'iiil 

5.26 

2.35 

0.47 

J.76 

188O0 

23500 

^JM  1(1(1(1 

3.02 

1.31 

0.33 

0.626 

1O5O0 

13100 

;»4  7<)0 

3.51 

1.78 

0.47 

(166 

1424C 

17800 

46300 

1.97 

1.15 

0.35 

I1..63 

920C 

11600 

28000 

2.20 

1.14 

0.46 

11.565 

91  IK 

11400 

241600 

7.30 

2.76 

0.67 

0.91 

2200f 

2/500 

98000 

4.25 

2.02 

0.47 

(1.73 

1616( 

20200 

,82400 

5.36 

2.36 

0.76 

(1.86 

1880f 

23500 

.30700 

2.53 

1.35 

0.51 

0.63 

10801 

13500 

93300 

1.32 

0.99 

0.41 

0.55 

792( 

9900 

61400 

4.00 

1.98 

0.48 

0.73 

16S4( 

19800 

50300 

3.53 

1.69 

0.48 

0.756 

13.50( 

16900 

,21000 

1.94 

0.98 

0.31 

0.58J 

784( 

9800 

,11200 

2.64 

1.46 

0.56 

0.76 

11701 

14600 

70600 

1.06 

0.71 

0.36 

0.50 

668< 

7100 

108300 

2.85 

1.53 

0.48 

0.68 

1224( 

1.6300 

73100 

l.Il 

0.62 

0.24 

(1.61 

490( 

6200 

77400 

1.96 

1.10 

0.64 

0.716 

880( 

11000 

60300 

0.75 

0.49 

0.30 

0.4  7( 

392( 

4900 

93300 

2.95 

1.43 

0.45 

0.781 

U44( 

14300 

78000 

1.84 

0.96 

0.34 

0.662 

760( 

9500 

72300 

2.12 

1.20 

0.47 

0.72.'- 

960( 

12000 

57300 

1.30 

0.80 

0.39 

0.63( 

640( 

8000 

34700 

0.62 

0.46 

0.28 

0.40( 

S68( 

4600 

53400 

1.50 

0.94 

0.36 

0.64( 

-621 

9400 

41200 

0.63 

0.44 

0.19 

0.49.' 

352( 

4400 

38000 

0.87 

0.68 

0.34 

0.61  ( 

544( 

6800 

25400 

0.43 

0.37 

0.25 

0.4  7( 

292f 

3660 

35000 

1.14 

0.83 

0.36 

0.68f 

664( 

8300 

27700 

0.79 

0.66 

0.33 

0.60t 

4512 

5640 

19500 

0.32 

0.31 

0.19 

0.46i: 

2481 

3100 

16200 

0.28 

0.22 

0.17 

0.38( 

176( 

2200 

18000 

0.47 

0.37 

0.19 

0.56S 

296C 

3700 

15100 

0.36 

0.33 

0.20 

().63(: 

2616 

8270 

18300 

0.29 

0.29 

0.19 

0.5  IC 

232C 

2900 

7100 

0.21 

0.24 

0.185 

(1.46(: 

192C 

2400 

4800 

0.08 

0.11 

0.096 

:).S7(1 

88C 

1100 

1900 

0.014 

0.032 

0.036 

0.200 

256 

360 

c 


Tl 


XXII. 

EVEN- 

-LEGGED 

ANGLES. 

THIS  TaBLB 

E,    SBB  TaBLB   XIX.) 

Ixis  Parallel  to  One  Side. 

M— tr-N 

Axis  ^ 

at  46°.  ^' 

1 

i 

05 

TrsujsverHe  Value 
(y)  iu  lbs. 

5i 
II'" 

For  Iron. 

For  Steel. 

i6 

7.54 

3.24 

1.860 

60350 

75400 

15 

1.37 

31 

7.01 

3.28 

1.820 

56100 

70100 

13.10 

1.34 

91 

4.61 

3.46 

1.685 

36900 

46100 

7.75 

1.35 

22 

3.90 

3.42 

1.580 

31200 

39000 

6.77 

1.35 

34 

4.97 

2.19 

1.010 

39760 

49700 

8.67 

.96 

70 

3.94 

2.37 

1.550 

31550 

39400 

6.07 

.98 

Jo 

2.64 

2.52 

1.460 

21100 

20400 

3.77 

1.02 

20 

3.28 

1.80 

1.396 

2G250 

32800 

4.88 

.772 

20 

2.24 

1.92 

1.286 

18000 

22400 

2.05 

.707 

5G 

2.47 

1.40 

1.271 

19800 

24700 

3.45 

.634 

18 

2.32 

1.39 

1 .  220 

18550 

23200 

3.01 

.624 

56 

1.52 

1.52 

1.138 

12200 

15200 

1.86 

,650 

J8 

2.40 

1.21 

1.220 

19200 

24000 

2.40 

.470 

58 

1.74 

1.08 

1.122 

13920 

17400 

2.04 

.470 

56 

1.15 

1.15 

1.013 

9200 

11500 

1.20 

.  184 

50 

.90 

1.12 

.930 

7200 

9000 

.95 

162 

n 

1.13 

.76 

.996 

9040 

11300 

1.05 

338 

13 

.96 

.79 

.930 

7080 

9000 

.95 

.336 

14: 

.58 

.86 

.842 

4640 

5800 

.52 

.361 

13 

.79 

.G6 

.887 

6320 

7900 

.61 

.300 

.5 

.59 

.71 

.802 

4720 

5900 

.50 

.309 

»5 

.48 

.72 

.780 

3840 

4800 

.39 

.303 

!2 

.61 

.52 

.770 

4880 

6100 

.52 

.221 

!8 

.66 

.57 

.806 

5280 

6600 

.51 

.227 

'0 

.39 

.59 

.717 

3120 

3900 

.30 

.252 

12 

.34 

.59 

.700 

2720 

3400 

.25 

.240 

!2 

.46 

.45 

.740 

3680 

4600 

.35 

.194 

'8 

.45 

.44 

.720 

3600 

4500 

.35 

.197 

iO 

.31 

.47 

.654 

2480 

3100 

.22 

.208 

:0 

.26 

.50 

.690 

2080 

2600 

.17 

.212 

■8 

.51 

.38 

.730 

4080 

5100 

.28 

.150 

:5 

.30 

.33 

.634 

2400 

3000 

.21 

.154 

11 

.22 

.33 

.580 

1765 

2200 

.13 

.158 

17 

.19 

.38 

.570 

1520 

1900 

.11 

.160 

8 

.35 

.31 

.640 

2800 

3500 

.18 

.120 

17 

.20 

.27 

.550 

1600 

2000 

.12 

.120 

8 

.15 

.29 

.507 

1200 

1500 

.08 

.129 

9 

.17 

.19 

.510 

1360 

1700 

.09 

.096 

6 

.14 

.19 

.487 

1120 

1400 

.07 

.083 

4 

.12 

.19 

.450 

960 

1200 

.06 

.081 

1 

.10 

.21 

.444 

800 

1000 

.05 

.094 

9 

.085 

.21 

.440 

680 

850 

.04 

.081 

123 

.126 

.152 

.400 

1010 

1260 

.05 

.070 

■77 

.079 

.138 

.404 

632 

790 

.04 

.071 

44 

.050 

.147 

.358 

400 

500 

.02 

.067 

37 

.047 

.084 

.340 

376 

470 

.02 

.045 

130 

.040 

.084 

.310 

320 

400 

.01 

.040 

f22 

.031 

.096 

.296 

250 

310 

.01 

.043 

19 

.029 

.066 

.286 

232 

290 

14 

.023 

.070 

.  264 

184 

230 

12 

.021 

.048 

.254 

168 

210 

1 

09 

.017 

.053 

.233 

136 

170 

"        1 

Table   XXIl. 
LIST   OF   IRON   AND   STEEL   EVEN-LEGGED  ANGLES. 

<FOa  INFOBUATIOK  A9  TO   USB  OF  THIS  TablB,   8BB  TaBLK   XIX.) 


13 

^■3 


C,E 

A,B,C,D,E.. 
A,b,C,D,E,F 

li,C 

C 

B,C,F 

U,U 

A 

A 

A,C,E,F.... 
B,A,C,E,r.. 
A,B,C,D,E,F 

E 

A,li.C,E,F.. 
A,B,C,D,E,F 

B 

A,C,E,F.... 
B,A,C,D,E,F 

A.B.C 

A.li.r.lM-  - 
A.I'.i  . 


6  xe 
6  xe 

6   X6 

6  XC 

5   X5 

X5 

X5 


X4 
XI 
3JX3t 
3JX3I 
3jx3j 
3iX3| 
3  X3 
3   X3 


01. 9 
37.5 
54.4 
51. S 


B. 

B,i;.i 

A,B,C,1>, 
L,B,C,D,E,F. 
I 

a,b,6,'d,e,f. 

A,B,U,D,E,F. 

C,B 

E,F 

A,B,C,D,E,F. 
A,B,C,D,E,F. 

C,E,F 

A,t:,E,F 

1,A,C,E.F... 
A,B,C,D,E,F. 

B,D 

P 

A.CE.F 

A,B,C,D,E,F. 

A,C,F 

B.A.CD.E.F. 
A,B,C,D,E,F. 

A,F 

A,F 

A,F 

A,F 


3.500 
3.438 
3.250 


2.781 
2.750 
2.500 
2.500 


1.C25 
1.504 
1.500 
1.500 
1.438 
1.375 
1.250 
1.125 
1.0C3 


35.46 
31.01 
19.91 
17.22 
19.04 
U.70 
9.33 


1.43 

2.17 
1.24 
1.025 
1.825 
1.405 

.72 
1.385 


5.44 
5.1C 
2.8C 
5.10 


2.81 
1.44 
2.77 


.145 
.100 
.125 
.085 


3.24 
3.28 
3. 40 


1.92 
1.40 
1.39 
1.52 


|4lll 

1.8G0 
1.820 
.685 
1.580 
I.GIO 
1.550 
1.4G0 
1.396 
1.286 
1.271 
1.230 
1.138 
1.220 
1.122 
1.013 
.930 
.996 


.56100 
36900 
31200 
39760 
31550 
21100 
26250 
18000 
19800 
18550 
12200 
19200 
13920 
9200 
7200 
9040 
7680 
4640 
6320 
4720 
3840 
4880 
6280 
3120 
2720 


2480 
2080 
4080 
2400 
1765 
1520 
2800 
1000 
1200 
1360 
1120 


75400 
70100 
40100 
39000 
49700 
39400 
26400 
32800 
22400 
24700 
23200 
15200 
24000 
17400 
11500 
SOOO 
11300 
9000 
6800 
7900 
5900 
4800 
6100 
6600 
3900 
3400 
4600 
4500 
3100 
2600 
5100 
3000 
2200 
1900 
3500 
2000 
1500 
1700 
1400 
1200 
1000 


1.37 
1,34 
1.35 


2.40 
2.04 
1.20 


r 


Table 
LND   STEEL  GLES. 

nON   AS    TO    USK   Oi 


■  Axis  Parai'l  to  a-b. 

Parallel  to  Short  I.ogto  Loii^ 

1-g.  , 

A-  .   .  .  - 

rt 

C<<'. 

<\-^. 

1 

.  .\>.  ,.^ 

o  »  ° 

1^  |I5'^ 

"So.  2 

S  J:.5'e 

Transver.ie  Value 
((■>  in  lbs. 

I'fc 

C  «  2 

For  Iron. 

For  Steel. 

9.72 

4.80 

2.   .77 

.96 

20880 

26100 

6.70 

0.71 

G.82 

4.88 

2.   .85 

.82 

15680 

19G00 

4.45 

0.72 

8.25 

4.08 

2.  1.17 

1.13 

25680 

32100 

8.35 

0.86 

6.87 

4.24 

2.  1.19 

1 .  03 

21600 

27000 

6.75 

0.90 

4.07 

4.33 

2.  1.32 

.91 

14000 

1 7500 

3.60 

0.88 

7 

3.42 

2.  1.19 

1.17 

25360 

31700 

7.46 

0.83 

6.75 

3.53 

2. '1.25 

1.08 

21450 

26800 

5.72 

0.83 

3.83 

3.70 

1.1. 34 

.96 

14720 

18400 

3.55 

0.85 

3.40 

3.72 

2  1.37 

.97 

13200 

16500 

2.89 

0 .  79  ■ 

6. Id 

3.46 

2.   .85 

1.01 

18880 

23600 

5.  75 

0.66 

4.!)2 

3.57 

2.   .92 

.90 

14480 

18100 

3.78 

0.67 

3.24 

3.72 

2.  1 

.82 

10100 

12600 

2.11 

0.62 

4.09 

2.99 

1.:  .94 

.91 

14000 

1  7500 

3.35 

0.64 

2.75 

3.13 

i.a.io 

.82 

9760 

12200 

2.14 

0.66 

4.G8 

2.28 

1.  1.28 

1.25 

24000 

30000 

6.10 

0.74 

3.49 

2.40 

l.<1.37 

1.11 

18480 

23100 

3.93 

0.74 

2.3;! 

2.53 

l.il.44 

1.04 

12400 

15500 

2.20 

0.69 

3.82 

2.39 

1.   .93 

.99 

14880 

18600 

3.72 

0.64 

2.30 

2.55 

1 . 1 1 . 05 

.86 

9680 

.12100 

1.96 

0.64 

2.03 

2.59 

1 . .  1 . 06 

.79 

8560 

10700 

1.94 

0.71 

3.72 

2.40 

l.f  .64 

.84 

11040 

13800 

2.58 

0.48 

3.32 

2.43 

l.f  .66 

.81 

10000 

12500 

2.11 

0.45 

1.88 

2.56 

l.j  .72 

.74 

6160 

7700 

.97 

0.41 

3.81 

2.10 

l.<  .77 

.94 

14800 

18500 

2.54 

0.48 

2.6G 

1.96 

1.;  .67 

.83 

9600 

12000 

2.45 

0.48 

1.83 

2.05 

1.-  .74 

.74 

7040 

8800 

1.25 

0.47 

3.08 

1.66 

1-1.21 

1.16 

20000 

25000 

2.80 

0.57 

1.99 

1.49 

1 . : 1 . 06 

.99 

12480 

15600 

2.29 

0.58 

1.43 

1.56 

i.n.io 

.89 

9120 

11400 

1.61 

0.61 

3 

1.69 

l.'^  .86 

.99 

14400 

18000 

2 .  05 

0.40 

1.97 

1  54 

1.;  .74 

.86 

9120 

11400 

1.56 

0.42 

1  .23 

1.61 

1.'-  .78 

.76 

5856 

7320 

.  77 

0.37 

2.40 

1.28 

1.1  .88 

1 .  04 

14400 

18000 

1.86 

0.39 

1.48 

1.14 

1.1  .76 

.88 

8640 

10800 

1.35 

0.40 

.74 

1.23 

l.<  .83 

.79 

4720 

5900 

.61 

0.39 

1.35 

1.21 

1.1  .52 

.72 

5760 

7200 

.80 

0.28 

.79 

1.25 

1.]  .53 

.62 

3440 

4300 

.43 

0.29 

.G9 

1.26 

1..^  .14 

.32 

1150 

1440 

.16 

0.134 

1.30 

1.14 

1.1  .36 

.61 

4480 

5600 

.80 

0.31 

.G3 

1.08 

1.1  .32 

.48 

2080 

2600 

.41 

0.32 

1.02 

.84 

l.<  .52 

.77 

5680 

7100 

.91 

0.33 

.56 

.89 

.S  .57 

.66 

3220 

4020 

.36 

0.28 

.87 

.86 

1 .  (  .28 

.58 

2920 

3650 

.47 

0.21 

.73 

.86 

l.(    .31 

.53 

2800 

3500 

.39 

0.22 

.48 

.93 

A    .U 

.48 

1840 

2300 

.24 

0.23 

.58 

.55 

.>■    .31 

.62 

3100 

3860 

.37 

0.18 

.4] 

.66 

.7  .35 

.54 

2030 

2530 

.  20 

0.18 

.31 

.40 

.!■  .16 

.44 

1350 

1680 

.15 

0.12 

.23 

.50 

.7  .18 

.38 

920 

1150 

.  08 

0.12 

.3G 

.42 

.7  .17 

.44 

1520 

1900 

.11 

0.095  • 

.  2;! 

.45 

.C  .15 

.  3  7 

960 

1200 

.08 

0.106 

.21 

.35 

.7  .17 

.  32 

712 

890 

.08 

0 .  nX4 

.IG 

.40 

.6  .15 

.31 

GOO 

750 

.05 

0.096 

Table  XXIII. 
LIST    OF   IRON   AND   STEEL   UNEVEN-LEGGED  ANGLES. 

(For  iNFORMATrox  as  to  Usr  of  this  Tadls,  asB  Taulk  XIX.) 


XXIV. 
VND   STEEL   TEES. 

E  OF  THIS  Table,  ske  Table  XIX.) 


to  Web.     /v\ ,^JL^j n 


gpo 


1.32 

1.42 

1.10 

.(i4 

.(;2 

.43 

.44 

.48 

1.14 

.77 

2.46 

1.93 

1.48 

1.10 

.76 

.49 

.28 

1.51 

1.12 

1.14 

.77 

1  .  54 

1.12 

1.10 

.80 

.83 

.82 

.58 

.52 

.53 

.31 

.32 

.12 

.17 

.40 

.30 

.86 

.71 

.61 

.57 

.58 

.10 

.47 


.99 

1.08 

1.05 
73 
.  77 
.61 
.58 
.66 

1.13 
.76 

1.57 


1.37 

1.18 

1.09 

.80 

.62 

.46 

1.24 

1.06 

1.03 

.85 

1  .  35 

1.15 

1.10 

.93 

.89 

.84 

.82 

.70 

.69 

.54 

.52 

.40 

.37 

.75 

.66 

.96 

.84 

.75 

.76 

.74 

.30 

.66 


Transverse  Value 
(i')  in  lbs. 


For  Iron.      For  Steel. 


16560 

16640 

17360 

8800 

8880 

6400 

5680 

6480 

17440 

6880 

24400 

19840 

15760 

15520 

7680 

4800 

2800 

14060 

11920 

10560 

8000 

16800 

13360 

10000 

8560 

6670 

6080 

6800 

4800 

4160 

2960 

2590 

1600 

1360 

5280 

4080 

6480 

5280 

6000 

4500 

3860 

800 

2820 


20700 

20800 

21700 

110(10 

1 1 1 00 

8000 

7100 

8100 

21.S00 

8600 

30500 

24800 

19700 

19400 

9600 

6000 

3500 

18700 

14900 

13200 

10000 

21000 

16700 

12500 

10700 

8340 

7600 

8500 

6000 

5200 

3700 

3240 

2000 

1700 

6600 

5100 

8100 

6600 

7500 

5630 

4830 

1000 

3520 


Axis  Parallel  to  Web.    ^\- 


-¥- 


A 


9.04 

5.25 

5.31 

5.70 

5 .  24 

5 .  23 

4.60 

3.94 

3.90 

2.39 

2.70 

2.70 

2.62 

3.23 

2 .  30 

2.00 

1.80 

1.80 

1.82 

1.53 

1.60 

1  .30 

1.40 

1  .06 

1.15 

.97 

.75 

.94 

.89 

.74 

.85 

.68 

.57 

.56 

.63 

.62 

.50 

.50 

.58 

.49 

.40 

.33 

.26 


3.01 
2.10 
2.12 
2.30 
2.10 
2.09 
1.80 
1.58 
1.70 
1.06 
1.40 
1.40 
1  .31 
1.61 
1.10 
1 

.91 
1 

1.04 
.87 
.  93 
.87 
.92 
.71 
.77 
.65 
.50 
.63 
.60 
.49 
.567 
.453 
.455 
.372 
.460 
.450 
.400 
.400 
.460 
.390 
.320 
.  260 
.231 


P5" 


1.46 
1.49 
1.49 
1.46 
1.37 
.86 
.96 
.64 
.67 
.70 
.77 
.81 
.86 
.92 
.52 
.56 
.53 
.59 
.36 
.38 
.37 
.42 
.46 
.38 
.40 
.44 
.42 
.49 
.47 
.38 
.50 
.30 
.34 
.26 
.28 
.30 
.28 
.27 
.37 


Transverse  Value 
(r)  in  lbs. 


For  Iron.     For  Steel 


24080 

16800 

16960 

18400 

16800 

16720 

14400 

12640 

13600 

8480 

11200 

11200 

10480 

12880 

8800 

8000 

7280 

8000 

8320 

6960 

7440 

6960 

7360 

5680 

6160 

5200 

4000 

5040 

4800 

3920 

4540 

3630 

3640 

2976 

3680 

3600 

3200 

3200 

3680 

3120 

2560 

2080 

1850 


30100 

21000 

21200 

23000 

21000 

20900 

18000 

15800 

1  7000 

10600 

14000 

14000 

13100 

16100 

lloOO 

10000 

9100 

lOOdO 

10400 

8700 

9300 

8700 

9200 

7100 

7700 

6500 

50O0 

6300 

6000 

4900 

5670 

4530 

4550 

3  720 

4600 

4500 

4000 

4000 

4600 

3900 

3200 

2600 

2310 


A,C,F 

1,B,D,E,K. 
A,C,E 


Talm.k    .WIV. 
LIST   OF   IRON   AND   STEEL   TEES. 

(Foe  Ikfobmation  as  to  thb  Usb  of  this  Table,  sbk  Table  XIX.) 


2.05 
1.75 
1.93 
1.03 


n.53 

)1.G4 
•  1.63 


3  1.50 
51.13 
J  1.03 
9  1.20 
U.21 
jl.02 
J  1.12 


50.944 
0.440 
0.376 


»  Normal  to  Web. 


2.08 
2.17 
l.IO 
1.11 


3.64 
3.20 
2.14 
5.55 


2. 38 
2. OS 
1.76 
1.73 
1.4G 
1.51 
1.12 
2.10 
1.88 
1.S5 
1.80 
1.9S 
1.73 
1.47 


l.(ie 
1.2G 
1.12 


1.93 
1.48 
1. 10 


1.51 
1.12 
1.14 


--.& 


1.57 
1.37 
1.18 


10560 
16G40 
1 7360 
6800 


24400 
19840 
157GO 
15520 
7680 
4800 
2800 
14900 
1 1 920 
10560 
8O00 
16800 
13360 
10000 
8560 
6670 
6080 
6800 
4800 
4100 
2900 
2590 
1600 
1360 
6280 
4080 
6480 
6280 
6000 
4600 
3860 


20800 
21700 
UOOO 
11100 
HOOO 
7100 
8100 
21800 
8600 
305O0 
24800 
19700 
19400 
9600 
6000 
3500 
18700 
14900 
13200 
10000 
21000 
16700 
12500 
10700 
8340 


8480 

10600 

2420 

3020 

1980 

2470 

1360 

1700 

430 

540 

2080 

2600 

1536 

1920 

1344 

1680 

1000 

1250 

--IH- 


ih 


2.30 
2.10 
2.09 
1.80 
1.58 
1.70 
1.06 
1.40 
1.40 
1.31 
1.61 
1. 10 


24080 
16800 
1 0900 
18400 
16800 
16720 
14400 
12640 
13600 
8430 
11200 
11200 
10480 
12880 
6800 
8000 


3010 
21000 
21200 
23000 
21000 
2O900 
18000 
16800 
17000 
10600 
14000 
14000 
13100 
16100 
llOOO 
10000 
9100 
lOOOO 
10400 
1700 


5680 
6160 
5200 
4000 
5040 
4800 
3920 
1540 
36S0 
3640 


3120 
2560 
2080 
1850 
1712 
1280 
1704 
1120 
1680 
1280 
1440 


7100 
7700 
6500 
5000 
6300 
6000 
4900 
6670 
4530 
4550 
3720 
4600 
4600 
4000 
4000 
4600 
3900 
3200 
2600 
2310 
2140 
1000 
2130 
1400 
2100 
1600 
1800 
1400 
1700 
12.50 
1140 


fA 


r^ 


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